For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

### Poisson Distribution

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Poisson Distribution
- Formula for the Poisson Distribution
- Key Properties of the Poisson Distribution
- Example I: Forest Fires
- Example II: Call Center, Part A
- Example II: Call Center, Part B
- Example III: Confirming that the Mean of the Poisson Distribution is λ
- Example IV: Find E (Y²) for the Poisson Distribution
- Example V: Earthquakes, Part A
- Example V: Earthquakes, Part B

- Intro 0:00
- Poisson Distribution 0:18
- Poisson Distribution: Definition
- Formula for the Poisson Distribution 2:16
- Fixed Parameter
- Formula for the Poisson Distribution
- Key Properties of the Poisson Distribution 5:30
- Mean
- Variance
- Standard Deviation
- Example I: Forest Fires 6:41
- Example II: Call Center, Part A 15:56
- Example II: Call Center, Part B 20:50
- Example III: Confirming that the Mean of the Poisson Distribution is λ 26:53
- Example IV: Find E (Y²) for the Poisson Distribution 35:24
- Example V: Earthquakes, Part A 37:57
- Example V: Earthquakes, Part B 44:02

### Introduction to Probability Online Course

### Transcription: Poisson Distribution

*Hi, these are the probability lectures here on www.educator.com, my name is Will Murray.*0000

*We have been working through the discrete distributions.*0004

*We are going to talk about the last distribution which is that the Poisson distribution.*0008

*After that, we will get into continuous distributions in some future lectures.*0013

*Let us jump right into Poisson.*0017

*The Poisson distribution describes events that occur randomly and independently.*0021

*The way we think about this is, every once in while something happens and*0027

*there is sort of no connection between different instances of this event.*0032

*The typical example of the Poisson distribution is, if you are working in a call center.*0037

*Maybe, you are working tech support for apple, or something like that.*0043

*You are just waiting for the phone ring, and every so often*0046

*you get a phone call from someone in the world asking you for help in some problem.*0050

*There is really no connection between the number of phone calls that you get from 1 hour to the next.*0056

*It might be that this hour you get 2 phone calls, in the hour you got 5 phone calls,*0062

*and the hour after that, you get no phone calls at all.*0068

*The random variable that we are keeping track of, is the number of calls that you get in an hour.*0071

*There is a lot of different physical phenomenon that can be modeled by the Poisson distribution.*0079

*Other typical examples would be, if you are sitting by the side of a country road and*0085

*just keeping track of the number of cars ago by per unit of time.*0090

*Maybe, each hour, 1 hour, there are 3 cars that go by.*0095

*In the next hour, there are 7 cars that go by.*0098

*In the next hour, there are 4 cars that go by.*0101

*There is no connection between one car and the next.*0104

*You are just keeping track of how many cars there are in any given unit of time.*0109

*Another example would be the number of earthquakes that strike a particular region per unit of time.*0115

*All these things are modeled by the Poisson distribution.*0121

*It counts the number of occurrences of a random event,*0124

*when there is no connection between one instance of the event and the next instance.*0128

*Let us go ahead and learn some formulas for the Poisson distribution.*0133

*There is one parameter that you have to keep track of, and it is traditional to use λ for this parameter.*0137

*That is the Greek letter λ right there.*0143

*Λ is the average number of occurrences of the event per unit of time, it does not have to be a whole number.*0146

*For example, if you are working in a call center and after many days, you kept track of the average number of calls.*0152

*On average, maybe you get 5 calls per hour would be λ.*0159

*It does not mean you are going to get 5 calls every hour.*0162

*Some hours you might get no calls at all.*0165

*Some hours you might get 15 calls, but that is the long-term average.*0166

*It is the only parameter that goes into the Poisson distribution is that value λ.*0172

*Like I said, it does not have to be a whole number.*0176

*The probability distribution is P of Y is equal to λ ⁺Y E ⁻λ/Y!.*0179

*That is for whole number values of Y.*0189

*It could be as little as 0, because if you are working in your call center and you are waiting for calls to come in,*0193

*it could be that you know get a whole hour and get no calls at all.*0202

*You could get one call, you could get 2 calls, you could get hundreds of thousands of calls,*0205

*if you are particularly unlucky that hour.*0211

*The range for Y is any whole number between 0 and infinity.*0213

*This formula for the Poisson distribution is a little difficult to memorize.*0222

*It is something you do need to memorize, if you are going to take probability.*0226

*I often remember it based on the Taylor series.*0230

*I remember the Taylor series for E ⁺λ.*0234

*You probably learn this using X, when you first learn about Taylor series nut I'm going to use it.*0238

*Let me remind you of the Taylor series for E ⁺X in the form that you probably know.*0247

*E ⁺X is the sum from N equals 0 to infinity of X ⁺N/N!.*0252

*We tend to use slightly different variables in probability, instead of N we use Y.*0261

*The X here is taking the place of λ, it is becoming λ.*0268

*If we change that according to those variables, then E ⁺λ is the sum from Y equals 0 to infinity.*0274

*X ⁺B becomes λ ⁺Y and N! becomes Y!.*0283

*If you look at that old formula for Taylor series, that is exactly the formula that we have for the Poisson probability distribution.*0287

*That is how you can remember that part, λ ⁺Y/Y!.*0303

*If you can remember the Taylor series for E ⁺λ.*0307

*That E ⁻λ, you want to think about that as a constant being multiplied on.*0311

*That E ⁻λ, λ is a fixed parameter, that is just a constant, that is independent of Y.*0316

*Just like every distribution, we have to figure out its mean, variance, and its standard deviation.*0326

*The key properties of the Poisson distribution here.*0335

*The mean is always the same as the expected value, those are synonymous.*0338

*The expected value and mean, those are always the same.*0342

*For the Poisson distribution, it is just λ by itself.*0346

*Of course, that should not be surprising, that should not be difficult to remember,*0350

*because we set this up knowing ahead of time that you average λ calls per hour.*0355

*The fact that the mean comes up to be λ is not all surprising, and in fact, it should look like that.*0362

*The slightly more surprising result is that, the variance also comes out to be λ.*0368

*Σ² here, the V of Y is also λ.*0374

*That is kind of nice to remember, that was not obvious.*0378

*If you do the calculus to calculate the variance, that it does come out to be λ.*0382

*Remember, the standard deviation is always the square root of the variance.*0388

*In this case, since, the variance is λ, the standard deviation is just the square root of λ.*0393

*Let us go ahead and talk about some examples of the Poisson distribution.*0401

*Our first example, we have California averaging 6 major forest fires per year.*0407

*Somebody has done a study, maybe over the last 50 years,*0412

*and figure out that on average California has 6 major forest fires per year.*0415

*They were interested in the chance that there will be exactly 4 fires in this coming year.*0420

*Also, what is the chance that there will be at least 4 fires?*0427

*That is a typical Poisson distribution problem because forest fires,*0432

*just every once in a while kind of essentially randomly, a forest fire happens.*0436

*There is not a lot of connection between one forest fire and the next.*0442

*Let me remind you of the basic formula for the Poisson probability distribution.*0447

*That was P of Y is equal to E ⁻λ × λ ⁺Y/Y!.*0452

*Remember, the λ use the parameter that represents the average number of occurrences per unit time.*0462

*In this case, we have been told that California averages 6 forest fires per year.*0469

*That is the value of λ that we are going to use, λ = 6.*0474

*The first question that we have to answer is, what the chance is that there will be exactly 4 fires this year?*0478

*That means, Y is equal to 4, we want to find the probability of 4.*0485

*That is E λ, -6.*0490

*Λ is 6, 6 ⁺Y is 6⁴, and Y! Is 4!.*0494

*This is a fairly amenable fraction to simplification.*0503

*Let me go ahead and simplify that a bit.*0508

*E⁻⁶, I will just put that in the denominator is E⁶.*0510

*4! Is 24, 6⁴ is 6 × 6 × 6 × 6.*0515

*I can simplify that a bit, I can cancel one of these 6 with the 24 and get 4 here.*0522

*4 is 2 powers of 2, I can cancel two of the 6 down to 3.*0529

*It is just taking some 2’s out of their.*0533

*I'm left with, in the denominator E⁶.*0536

*In the numerator, 3 × 3 × 6 is 9 × 6 = 54.*0540

*That is my probability of there will be an exactly 4 fires in the next year.*0549

*I could also calculate a decimal approximation of that and I did that with my calculator.*0558

*I just did 54 divided by E⁶ and it came out to be about 13.4%.*0564

*If somebody wants to know, what is the chance of getting exactly 4 forest fires next year,*0571

*it will come out to be exactly 13.4%, or approximately 13.4%.*0577

*The second part of this question is, what is the chance that there will be at least 4 fires next year?*0584

*That is the probability that Y is greater than or equal to 4.*0592

*I think the easier way to think about that is to reverse the question and say,*0598

*what is the probability that there will be 3 or fewer fires?*0603

*Because we can just add up from 0 to 3, and then subtract that from 1.*0607

*This is 1 -, the probabilities of their being fewer fires, there could be no fires at all.*0611

*Also, P of 0, let me put this in parenthesis.*0618

*P of 0 + P of 1 + P of 2 + P of 3, the probability that there will be anywhere from 0 through 3 fires, next year.*0623

*That is the opposite of what we are calculating.*0635

*We can use that to find the probability that there will be 4 or more fires.*0638

*We are going to use this formula, this generic formula for the Poisson distribution.*0643

*Notice that, there is E ⁺λ and that is independent of Y.*0648

*I’m going to factor that out.*0653

*This is 1 – E⁻⁶.*0655

*The other terms are just λ ⁺Y/Y!.*0663

*In this case, for Y = 0, 6⁰/0! + 6¹/1! + 6²/2! + 6³/3!.*0669

*And that, we will simplify quite a bit, 6⁰/0!, that is 1 + 6 + 6²/2 + 6³/3! Is 6.*0688

*By the way, what you should notice here is that, we are essentially finding the first few terms of the Taylor series for E⁶.*0705

*This is really the Taylor series for E⁶, 1 + 6 + 6²/2! + 6³/3!.*0715

*If you recognize that, that can really help you check your work when you do a Poisson distribution problem,*0733

*because the Poisson distribution is very closely connected to the Taylor series for E ⁺X.*0738

*Let me simplify this with the fractions work out fairly nicely here.*0747

*I will put the E⁻⁶ as the denominator.*0753

*I think it will work a little more nicely there, E⁶ in the denominator, that is 7 +, 6² is 36 divided by 2 is 18.*0756

*6³ divided by 6, we would have a cancellation of a 6 from both top and bottom there.*0767

*We need to calculate 6³, it is just 6².*0774

*6² is 36.*0779

*What we get there is 1 -, 7 + 18 is 25 + 36 is 61.*0783

*1 - 61/E⁶ is our exact probability there.*0792

*I did calculate that into a decimal.*0798

*My decimal approximation, when I put that out in my calculator was 84.9%.*0802

*84.9% is approximately the probability that there will be at least 4 forest fires in California next year.*0810

*It is pretty high there, I do expect to see at least 4 forest fires in California, probably more.*0820

*That is not surprising, since the average number of fires per year is 6.*0828

*Let me recap where everything came from here.*0832

*It came from this master Poisson probability distribution formula P of Y is E λ × λ ⁺Y/Y!.*0835

*The λ here is the average number of occurrences, which in this case was 6, 6 fires per year, that is where we got the λ.*0846

*We plug that 6 in here for λ everywhere, and then it is a question of figuring out what values of Y you are interested in.*0856

*First, we are figuring out exactly 4 fires.*0864

*That is why I plugged in Y is equal to 4.*0867

*I plugged that into the formula, simplify it down, and I got the probability to be about 13.4% .*0871

*At least 4 fires means the probability that Y will be greater than or equal to 4, that is hard to calculate directly.*0879

*Instead, I flipped it around and calculated the probability that Y will be less than or equal to 3.*0887

*That is what I'm doing here, P of 0, P of 1, P of 2, P of 3,*0893

*that is the probability that you will have no fires, or 1 fire, or 2 fires, or 3 fires.*0897

*And then, I subtracted that from 1 to get the probability of being greater than 4.*0903

*For each one of these, I used this Poisson distribution formula again.*0908

*Since, there was E ⁺λ in all of these and that is a constant, I just factor that out.*0915

*That is what this E⁻⁶ is doing out here.*0920

*And then I went through the other values of Y and drop them in, 0, 1, 2, and 3.*0923

*It looks like I forgot the point on my factorial there.*0929

*I drop those values in and what I noticed was that this turns into the Taylor series for E⁶.*0935

*That helps me check my work a little bit.*0943

*It does not really how we calculate it but it helps me check my work.*0945

*And then, I just simplify the fractions and reduced it down to a percent, 84.9%.*0948

*On our second example, we have a call center receiving 2 calls per minute, on average.*0957

*At first, we are going to use Markov’s inequality to estimate the chance that fewer than 5 calls will come in, in the next minutes.*0965

*It looks a little strange, let me write that down because the word in, occurs twice.*0975

*That is saying 5 calls will come in, in the next minute.*0978

*In the second part, we are going to find the exact chance that fewer than 5 calls will come in, in the next minute.*0983

*Let me remind you what Markov’s inequality was.*0991

*That was something that we learn several lessons ago, there is a video here on Markov’s inequality.*0994

*If you do not remember that, just scroll up and you will see a whole video on Markov's inequality, that can help you out there.*1000

*I will give you the short version of that, right now.*1006

*It says that, let me go ahead and label this.*1011

*This is for Markov’s inequality, says that the probability that Y is bigger than or equal*1015

*to some constant A is less than or equal to the expected value of Y divided by that value of A.*1025

*In this case, we are interested in the probability that fewer than 5 calls will come in.*1035

*That means A is equal to 5, the probability that Y is greater than or equal to 5.*1041

*The problem actually asked for fewer than 5, we will turn this around in a moment.*1046

*I want to calculate this up.*1051

*E of Y, that is the average value, the expected value, the mean.*1053

*In this case, that is μ is equal to 2 because we have been told that this call center averages 2 calls per minute.*1058

*This is less than or equal to 2/, in this case, 5.*1068

*If we turn that around, if we found that the probability that Y is greater than 5 is less than 2/5.*1076

*The probability that Y is less than 5 is then, if we flip that around must be greater than 1 - 2/5 which is 3/5.*1085

*If you convert that to a percentage is 60%.*1100

*Our answer here is, the probability that Y is less than 5 is greater than 60%.*1104

*That is what Markov’s inequality told us.*1115

*I'm going to go ahead and answer the second part of this problem on the next slide.*1118

*Let me recap what we did with this part of the problem.*1123

*First of all, I did not use anything to do with the Poisson distribution here.*1126

*There is Poisson in this first part of the problem, I was just using Markov’s inequality.*1131

*If you want a refresher course in Markov’s inequality, we got a whole video on Markov’s inequality in this series.*1136

*Just go back and check out that video Markov’s inequality and you can get plenty of practice with that.*1143

*The short version of that is this formula.*1148

*The probability that Y is bigger than any constant A is less than or equal to the expected value of Y divided by A.*1151

*In this case, our A is this 5 right here.*1161

*The expected value of Y is the average number of calls per minute.*1165

*That is E of Y right there, we are given that in the problem.*1169

*That is where we get the value of 2 and the value of 5, in my answer here.*1174

*But what Markov’s inequality tells us is, it gives us a bound for the probability of Y being greater than 5.*1180

*In order to find the probability of Y being less than 5, you have to reverse that and reverse the inequality.*1188

*Instead of 2/5, we get 1 -2/5.*1198

*Instead of less than or equal to, it turns into greater than.*1200

*1 – 2/5 simplifies down to 3/5 or 60%.*1205

*Our answer is really the probability that Y is less than 5 is greater than 60%.*1209

*It is at least 60%.*1216

*If I'm making my plans for my call center and maybe I only have 5 lines on my telephone,*1219

*and I’m worried that maybe I will get more than 5 calls per minute.*1226

*I can say for sure that the probability of Y being less than 5 is at least 60%.*1231

*I got a 60% chance that I would not over load the 5 lines on my telephone.*1236

*There is no Poisson distribution coming in yet, that is coming in the answer to part B.*1241

*I will go ahead and jump to a new slide to do that.*1246

*In example 2, we have already found the probability of Y being less than 5 using Markov,*1251

*but we also want to calculate exactly the probability of Y being less than 5.*1259

*Now, we are going to use the Poisson distribution formula for that.*1263

*Let me remind you what the Poisson distribution formula is.*1270

*P of Y is equal to E λ, that is the constant part, Λ ⁺Y/Y!.*1273

*In this case, we want the probability that Y is less than 5.*1284

*In other words, we are going to get fewer than 5 calls that mean 0 through 4 calls.*1287

*The probability that Y is less than 5, we can get that by adding up the probabilities of getting 0 calls,*1294

*the probability of getting 1 call, 2 calls, or 3 calls, or 4 calls, + P of 3 + P of 4.*1301

*We have been told that our average number of calls per minute is 2.*1311

*I’m going to fill in λ is equal to 2 everywhere.*1315

*And that E λ, that is a constant, it does not depend on Y.*1319

*I will factor that out.*1323

*This is E⁻².*1324

*I’m going to plug in the values of Y being 0, 1, 2, 3, and 4.*1328

*That is 2⁰/0! + 2¹/1! + 2²/2! + 2³/3! + 2⁴/4!.*1335

*Let me go ahead and make my denominator E².*1356

*2⁰/0! Is 1, + 2, + I’m going to keep that expanded 2²/2! + 2³/3! + 2⁴/4!.*1360

*The reason I left that a little bit expanded was to kind of remind you that,*1378

*the Poisson distribution is related to the Taylor series expansion of E ⁺λ.*1385

*What we are really have here is the beginning of the Taylor series, the Taylor polynomial, for E ⁺λ, in this case, that is E².*1391

*1 + 2 + 2²/2! + 2³/3!, 2⁴/4!.*1405

*That helps you check that you have done your arithmetic right,*1412

*if it starts to look like the Taylor series for E ⁺X or E², in this case.*1415

*Now, let me start simplifying this, this is 3 + 2²/2! That is 4/2 is 2.*1421

*2³/3! Is 8/6 +⁴ is 16, 4! Is 24/E², that is 5 + 8/6 is 4/3 + 16/24 is 2/3 /E².*1428

*5 + 4/3 + 2/3 is 6/3 that is 2/E².*1453

*This is 7/E² which I did some calculations that I have written down somewhere.*1459

*7/E² on my calculator simplify down to 94.7%.*1466

*That means, it is very likely that you will have fewer than 5 calls.*1474

*If you are worried about getting more than 5 calls in your call center, maybe you have 5 lines on your telephone,*1480

*you can rest assured that with 95% chance, you will have fewer than 5 calls.*1489

*Notice that, with a Markov estimation, we got that the probability was greater than 60%.*1495

*That was the answer we got, that was on the previous slide.*1504

*You can scroll back and see that, if you do not remember how we got that.*1506

*It did not say it is equal to 60%, it said that the probability was greater than 60%.*1510

*This checks that because 94% certainly is greater than 60%, but it is a much more precise answer.*1518

*Markov’s inequality just gives you a very rough check, something you can calculate easily and quickly,*1525

*but it would not be the most accurate estimate.*1532

*Using the Poisson distribution, we got an exact answer at the cost of having to do a few more calculations.*1536

*To remind you how we got those calculations, we use this Poisson formula P of Y = E ⁺λ × λ ⁺Y/Y!.*1543

*Λ was 2, that is because the average number of calls per hour is 2, that was given in the stem of the problem.*1554

*I dropped λ = 2 in, everywhere.*1560

*To find the values of Y, I want all the values of Y less than 5 including 0.*1564

*I ran through and I plot all these values of Y into that formula and I factored out the E⁻²*1569

*because that was constant for all of them, that do not depend on Y.*1577

*Then, it was just a matter of simplifying down the fractions and simplify down to 7/E², and that converted to 94.7%.*1580

*Along the way, I have this check that the expansion I got looked exactly like the Taylor series expansion for E ⁺X, if X is equal to 2.*1590

*That is my exact probability of getting fewer than 5 calls,*1601

*of getting anywhere from 0 through 4 calls in our call center in the next hour.*1606

*Example 3, we are going to use the definition of expected value to confirm that the mean of the Poisson distribution is λ.*1615

*I told you at the beginning of this particular video that the mean is λ.*1624

*That is not in doubt, if you trust me there.*1630

*In example 3, we are actually figured that out from scratch using the definition of expected value.*1632

*Let me remind you what that definition was.*1639

*The expected value of a discreet random variable which Poisson distribution is discreet, it is not continuous.*1641

*It is the sum overall possible values of Y of Y × the probability of that particular value of Y.*1650

*In this case, the values that a Poisson variable can take are 0 to infinity.*1660

*In this case, we have the sum from Y equal 0 to infinity of Y.*1669

*The Poisson distribution formula is the E ⁻λ × λ ⁺Y/Y!.*1676

*Let me go ahead and factor out the E λ there because that is constant, that does not depend on Y.*1686

*We get E λ × the sum from Y = 0 to infinity of Y × λ ⁺Y/Y!.*1693

*Now, that looks a lot like the Taylor series for E ⁺X.*1704

*Let me remind you what the Taylor series for E ⁺X is.*1709

*E ⁺X is the sum from N equals 0 to infinity of X ⁺N/N!.*1712

*What we have here is something very like that.*1720

*The only difference here, it is different variables but that is no big deal.*1723

*The difference here is that, there is this extra multiple of Y on the outside.*1727

*We have to figure out a good way to handle that.*1733

*Let me show you how I can think about this.*1735

*I'm going to define F of λ to be E ⁺λ.*1737

*I'm going to expand that into a Taylor series because basically,*1745

*the reason I'm doing that is because I see here that there is a λ ⁺Y.*1751

*The λ is taking the place of the X, in our original Taylor series.*1755

*I will expand that it out into a Taylor series, E ⁺λ is the sum, I will put it in terms of Y.*1759

*Y = 0 to infinity of λ ⁺Y/Y!.*1767

*That looks a bit like what we are looking for but it does not have this extra factor of Y.*1776

*Let me show you how we can get that.*1781

*What we will do is, we will take the directive with respect to λ, take derivative of λ.*1783

*We will get F prime of λ.*1792

*The derivative of E ⁺X is just E ⁺X.*1797

*The derivative of E ⁺λ is just E ⁺λ.*1800

*It looks like I did not do anything there but I really did d by d λ.*1803

*I just did not change the function.*1807

*Now, I have to take the derivative of this series and this where you have to remember,*1809

*we are taking the derivative with respect to λ.*1816

*The Y! Is just a constant but the derivative of λ ⁺Y, we can use the power rule into Y × λ ⁺Y-1.*1819

*That is using the power rule there from calculus.*1832

*It is just a little strange because you are probably not used to using λ as your variable.*1838

*You are not used to having Y in the exponent, instead of N.*1843

*It is still the same rules from calculus.*1848

*I see that I have got something that is a little bit closer to the series that I'm looking for.*1852

*I got the factor of Y which is very keys.*1858

*I lost one of my λ because I have λ¹ -1, instead of λ ⁺Y.*1860

*I’m going to bump it up by another power of λl.*1868

*I will multiply by λ to bump it up.*1870

*I got λ × prime of λ, which is λl × E B ⁺λ.*1878

*I’m going to bring this λ inside, the sum for Y = 0 to infinity of Y.*1887

*Since, I have one more λ now, I get λ ⁺Y instead of Y -1 Y! in the denominator.*1892

*Now, I essentially have this series that I was looking for.*1901

*I got the right series, just need that factor of E ⁻λ on the outside.*1905

*I will multiply by E λ.*1911

*E λ × prime of λ is equal to E λ × λ × E ⁺λ, which is equal to E λ × the sum from Y = 0 to infinity of Y × λ ⁺Y/Y!.*1923

*This perfect because, on the right hand side, I got exactly that expression that I started out with up here.*1947

*That is E of Y right there.*1955

*On the middle term, this λ and this E ⁺λ, and E λ, those cancel each other out.*1958

*That is just E ⁺λ divided by E ⁺λ.*1966

*Those cancel each other out and I just get λ there.*1971

*That is exactly what I wanted, I wanted to show, remember that the expected value*1975

*or the mean of the Poisson variable is λ.*1980

*That is what I have done.*1984

*You can really trust that, you do not have to take my word from the initial slide.*1986

*You can really understand yourself why the mean comes out to be λ.*1992

*Let me remind you of the stuff you went through there.*1997

*I use the definition of expected value here, the sum on Y of Y × P of Y.*1999

*That was our definition of the mean or the expected value of a discreet random variable.*2005

*I expanded that out, the P of Y, I just remember the formula for the Poisson distribution.*2013

*I will plug that in.*2019

*The E λ was a constant, I pull that out here.*2021

*I have this kind of complicated series and I did not really know what to do.*2026

*I went back and look to my E ⁺X formula.*2030

*I was trying to sort of build up this series here.*2033

*I built it up step by step, I start out with E ⁺λ.*2036

*I expanded that out as a Taylor series.*2041

*I took its derivative, the reason I did that was to get that extra power of Y on the outside,*2045

*because I know that I can produce that using the power rule.*2052

*I need to build that because I had that inside the original series here.*2055

*That worked pretty nicely, if we are using the power of Y but then I lost the power of λ from the exponent there.*2063

*I multiplied it back on that power of λ and that bumped the λ ⁺Y -1 backup to λ ⁺Y.*2069

*That pretty much matches my series, except for this E λ.*2079

*I multiplied it by E λ everywhere, there it is.*2083

*That gave me the exact series that I wanted for E of Y.*2088

*Over here, the E λ and E ⁺λ, those cancel each other out and I just simplified down the λ.*2092

*We get that the mean that the Poisson random variable is λ.*2099

*This is not surprising at all, because we started out by assuming that we knew the average number of calls per hour.*2103

*We assumed that the average was λ.*2111

*This is not surprising but it just kind of checks all our assumptions here, and it checks our arithmetic.*2113

*Let us keep going with this, for example 4.*2122

*In example e, we have been asked to find E of the expected value of Y² for the Poisson distribution.*2126

*Now this time, it does not say we have to calculate that from scratch.*2132

*I'm going to go ahead and use some of the information from some earlier slides.*2135

*In particular was, I know that the expected value of the Poisson distribution is λ.*2141

*I also know that the variance σ² V of Y, which is the same as σ²,*2148

*those are 2 different notations for the same thing.*2155

*It is the expected value of (Y² - the expected value of Y)².*2158

*That is true for any distribution not just Poisson.*2166

*It is a very useful thing to remember.*2168

*For the Poisson distribution, the variance is equal to λ.*2171

*We were given on the third slide of these videos,*2176

*if you scroll back to the key properties slide, you will see that the variance is λ.*2181

*Let me fill in what I know here.*2188

*The E of Y² then, is equal to λ + E (Y)², that is λ.*2190

*I already said that the expected value of Y is λ.*2203

*That is λ + λ² there.*2206

*That is what we are supposed to find out.*2209

*We are going to be using this, by the way, in the next examples.*2211

*I want you to hang onto this and remember that it will be useful for the next example.*2215

*But in the meantime, let me just remind you of those steps there.*2220

*I’m really using the basic key properties of the Poisson distribution.*2223

*The expected value of Y is λ and the variance is also λ.*2227

*But also remember that, for any distribution, the variance is always equal to the expected value of Y² - the expected value of (Y)².*2237

*I plugged that in and I know that the expected value of (Y)², I will plug that in here to get λ².*2248

*I get the expected value of Y² is λ + λ².*2261

*Hang onto this, we will be using it in the next example, in example 5.*2265

*We got something cooked up that is going to require us to use the expected value of Y².*2271

*In example 5 here, we got California averages 2 major earthquakes per decade.*2279

*Y is going to represent the number of major earthquakes in the next decade.*2287

*We got a cost function here, the cost of damages, it depends on how many earthquakes you have.*2292

*In this case, the cost is 2Y² + 5Y + 10.*2299

*We want to find the expected cost, how much do we expect on average to pay in damages per decade,*2306

*for earthquakes in California.*2313

*We also want to find the probability that the damages will cost more than $40,000,000.*2315

*What is the probability next year that we will have to spend more than $40,000,000 on earthquake damages?*2321

*I guess not next year but next decade.*2329

*Let us find the expected cost first.*2331

*The expected cost here, the expected value of the cost is the expected value.*2336

*What we are given that the cost is 2Y² + 5Y + 10.*2342

*What we are really going to use very heavily here is linearity of expectation.*2350

*Expectation is always linear so I can expand this out into 2 × the expected value of Y².*2357

*I cannot bring that square out but I can bring the 2 outside, and I can separate out the different terms.*2370

*5 × the expected value of Y + the expected value of 10.*2375

*The expected value of 10 is just 10, there is not much to say there.*2381

*2 × the expected value of Y².*2386

*I figured out in example 4, that the expected value of Y² was λ² + λ.*2389

*That was in example 4, if you did not just watch it then not you want to go back and look at example 4,*2398

*and see how we figure that out.*2404

*I’m not going to recalculate that now, but you can go back and check the earlier part of the video on example 4,*2406

*if you have not just watched it.*2416

*This is 2 × λ² + λ + 5 × expected value of Y.*2418

*Since, this is a Poisson distribution, that is 5 × λ + 10.*2427

*This is 2 λ² + 2 λ + 5 λ is 7 λ + 7 λ + 10.*2433

*Let me go ahead and work that out with λ = 2.*2453

*The reason I'm using λ is equal to 2 is because we have been told that California averages 2 major earthquakes per decade.*2459

*That is λ is equal to 2 here.*2468

*This is 2 × λ² is 2 × 4 that is 8 + 7 × 2 is 14 + 10.*2474

*8 + 14 is 22, that is 32.*2489

*That tells me that my expected cost, the expected amount that I should average on*2500

*spending on damages per earthquake per decade is, I guess this is in millions of dollars.*2510

*This is $32,000,000, that means in long-term California can expect*2515

*to spend an average of $32,000,000 on earthquake damage each decade.*2526

*If you want to make some planning based on how much it would cost to retrofit buildings,*2533

*or how much it would cost to make up different earthquake safety plans,*2542

*you want a budget that against the expected amount of damages which is going to be $32,000,000.*2546

*Let me do part B of this problem on the next slide here.*2553

*Let us just quickly check here that the expected costs here was,*2558

*the expected value of what we are given the cost functioned 2Y² + 5Y + 10.*2565

*We expanded that out using linearity of expectation.*2576

*It is very useful to use linearity of expectation.*2579

*We have the expected value of Y² was λ² + λ.*2583

*That came from example 4, you can go back and watch the video in example 4, if do not know where that came from.*2587

*The expected value of Y is just λ, we have proved it in examples 3.*2593

*I also gave you the answer back in one of the first slides in this lecture.*2600

*Of course, the expected value of a constant 10 is just 10.*2605

*Then, I expanded that out and I got 8 + 14 + 10.*2610

*That was using the value of λ = 2 that I got from the stem of the problem, that we are averaging two earthquakes per decade.*2615

*I gave you 32 as an answer.*2625

*Apparently, our units here are millions of dollars.*2628

*We are talking about $32,000,000 for our expected cost for earthquake damage next decade.*2630

*Let us go ahead and jump to the next slide.*2638

*We will find the probability that the damages will cost more than $40,000,000.*2640

*That is what we are going to calculate here on example 5, the probability that the damages will cost more than $40,000,000.*2647

*We already found the expected cost on the previous slides.*2655

*We are already done with one.*2658

*In this case, the cost function here, the probability that C is greater than 40.*2660

*Let me write that as C is greater than 40.*2669

*I want to convert that into values for Y, let us see how that works out.*2672

*The expected cost was 2Y² + 5Y + 10.*2677

*I want to see if that is greater than or equal to 40.*2685

*Let me try and simplify this as much as I can but it is not going to simplify terribly well.*2689

*Let me subtract 10 from all sides, 2Y² + 5Y.*2693

*I forgot my 5 there.*2699

*5Y is greater than or equal to 30.*2701

*I want to figure out which values of Y I will be looking at there, and then I can find the probabilities.*2706

*Remember that Y is a whole number because with Poisson random variables, you are always getting whole numbers.*2712

*In this case, we are counting the number of earthquakes we might have in the next decade.*2720

*We are going to have a whole number of earthquakes.*2723

*You cannot have half an earthquake.*2726

*In this case, let us figure out which values of Y might give us 2Y² + 5Y greater than or equal to 30.*2729

*There is not a good way to solve that for whole numbers.*2737

*For something better, I will use trial and error.*2742

*Let me just plug in some values of Y.*2748

*If Y equals 0, we will get 2Y² + 5Y is 0 which is not greater than 30.*2750

*That certainly does not work there.*2757

*Let me try Y is equal 1, we will get 2 + 5, is that greater than 30?*2761

*No, that certainly is not.*2769

*Let us try Y equals 2, 2Y² is 2 × 4 that is 8 + 5Y is 10.*2774

*Is that greater than 30? No, that is not greater than 30 because that is 18.*2782

*Let us try Y equals 3, 2Y² is 2 × 9 is 18 + 5 × Y is 15, 18 + 15 is that greater than 30?*2788

*Yes, that is greater than 30 because that is 33.*2802

*Y = 3 or higher, essentially, if we have 3 or more earthquakes next decade then our cost goes above $40,000,000.*2806

*We want to find the probability of having 3 or more earthquakes next decade.*2817

*Let me try and calculate that out.*2825

*The probability that Y is greater than or equal to 3.*2828

*If we have 3 more earthquakes, we bust the budget.*2830

*If we can have fewer than 3 earthquakes, then we will stay under budget.*2833

*Assuming, we budgeted $40,000,000 for earthquake repair.*2837

*It is hard to calculate P of Y being greater than or equal of 3.*2842

*What is easier to calculate is to find the probability of it being less than 3.*2846

*Let me calculate the probability of 0, 1, 2, earthquakes and*2851

*then I will subtract that off from 1 to get the probability of their being more than 3 earthquakes.*2859

*I should have written p of 2 there, that is 2.*2865

*Now, I have to calculate these probabilities.*2873

*Let me remind you of the Poisson distribution formula.*2875

*The Poisson distribution formula tells me that P of Y is equal to E λ × λ ⁺Y/Y!.*2879

*In this case, our λ = 2.*2893

*That E λ is going to be constant for all these terms.*2898

*This is 1 – E⁻², that is the constant part.*2901

*I’m going to fill in the different values of Y with 2 being λ.*2907

*2⁰/0! + 2¹/1! + 2²/2!.*2911

*This is 1 -, I will E² as a denominator there.*2922

*2⁰/0! Is 1 + 2 + 4/2 is 2.*2927

*I get here 1 -5/E², that is my exact probability that there will be more than 3 earthquakes,*2935

*which means we will bust our budgets of $40,000,000 for damages.*2944

*I did throw that into a calculator, I found the decimal.*2950

*It worked out to about 32.3%.*2953

*If I have budgeted $40,000,000 for earthquake repair in California,*2959

*and I'm worried about be the prospect of going over budget, I should be worried because there is a 30% chance,*2965

*actually a little more than 30% chance that I will go over budget.*2973

*That I will have more than 3 earthquakes which mean that I will be spending more than $40,000,000 on earthquake repair.*2977

*Let me recap the steps here.*2987

*We are interested in whether C is bigger than or equal to 40.*2989

*I plug in my formula for C here, there is my cost of being bigger than or equal to 40.*2993

*That simplify down a little bit to Y² + 5Y is greater than or equal to 30.*2999

*I was trying to figure out which values of Y would make that true, that is because Y is a discreet random variable.*3004

*We are not looking for a real number for Y, we are looking for whole numbers for Y.*3011

*Since, there is not an easy way to solve that, I just use trial and error.*3018

*I plugged in some different values of Y until I figure out which values of Y would make that true.*3022

*I will plug in these Y’s in here.*3027

*The first one that worked was Y = 3.*3030

*Of course, all the other Y after that will make it true.*3035

*We are trying to calculate the probability that Y is bigger than or equal to 3.*3039

*That is not an easy thing to do directly with the Poisson distribution.*3044

*Instead, what you want to do is calculate the probability that Y is less than 3,*3049

*which means the probability of getting 0, 1, or 2, and then you subtract it from 1.*3057

*I used my Poisson distribution formula.*3064

*In particular, I notice that one of these factors is constant.*3068

*I factor that out right here, that is the E⁻².*3071

*The other terms, I was just plugging in the different values of Y.*3075

*Y= 0, 1, 2.*3080

*All those gave me pretty simple numbers to simplify down.*3083

*That is my exact answer, 1 - 5/E².*3087

*I got a decimal approximation as 32.3%, that is the probability that*3091

*there will be more than 3 or more earthquakes 3 in California next year.*3098

*In turn, that tells me the probability my might cost is going to go above $40,000,000 in damages next year.*3104

*That wraps up our lecture here on the Poisson distribution and that also wraps up our chapter on discrete probability distributions.*3113

*The next chapter is on continuous probability distributions.*3122

*We will be talking about intervals, instead of summations.*3126

*That will be a lot of fun and I hope you will stick around for that.*3129

*In the meantime, you are watching the probability lecture series here on www.educator.com.*3132

*My name is Will Murray, thank you for joining me, bye.*3137

2 answers

Last reply by: Dr. William Murray

Sat Feb 27, 2016 10:30 AM

Post by YILEI GE on February 24, 2016

Hi, am i wrong if I use Markov's ineuality for example 5?

1 answer

Last reply by: Dr. William Murray

Fri May 30, 2014 4:04 PM

Post by Carl Scaglione on May 29, 2014

Dr. Murray,

On this slide, referring to the last entered equation, I see the following:

e^(-lambda) * lambda * f^prime (lambda) = ... .

In your entry, lambda is missing which was multiplied through from the previous equation, but it requires an explanation. I do not see that it influences the outcome, i.e. Expected value = lambda, but its absence is notable.

1 answer

Last reply by: Dr. William Murray

Mon Mar 31, 2014 10:57 PM

Post by Burhan Akram on March 27, 2014

In the last example, You could have easily solve it through quadratic formula and you would get around 3 :)