Mean-Variance Ratio of the Poisson DistributionDate: 03/27/2001 at 19:08:10 From: Jocelyn Subject: Proof of mean/variance = 1 for Poisson I would like to show that V(X)/E(X) = 1 for the Poisson distribution. In other words, the ratio of the variance and the expected mean equals one. I know the formula for the Poisson distribution and I think that E(X) = x * f(x) and V(X) = x^2 * f(x) but I am not sure about that. In order to show this ratio I need to do a derivative under a sum, which I am not sure how to do either. Any help would be appreciated. Thank you. Date: 03/27/2001 at 21:58:43 From: Doctor Jordi Subject: Re: Proof of mean/variance = 1 for Poisson Hello, Jocelyn - thanks for writing to Ask Dr. Math. Yes, you are on the right track for finding the mean and variance of the Poisson distribution. Doing the summations (with the help of a few clever tricks) will yield the desired results. I can show you how to do it. First, let's make sure you and I agree. The probability density function of the Poisson distribution is given by: [L^x]*[e^(-L)] p(X = x) = ---------------- x! Where I have used capital L to represent the parameter of the distribution. Traditionally, the Greek letter Lambda is used for this parameter. I will keep calling it L from now on, though. Finding E(x) = mean of the Poisson is actually fairly simple. We go back to the definition of E(x) to see that we need to find the following infinite sum. inf --- \ [L^x]*[e^(-L)] E(X) = / x * ---------------- --- x! x=0 We will apply a standard procedure for summing up this series or a very often-used trick, if you like. (My professors like to say that "a twice-used trick is a procedure.") We will factor out whatever we can from under the sigma sign, trying to arrive at something like (Some Expression) * Infinite Sum of p(x) And since we already know that p(x) is the density function of the Poisson, we know that it must sum up to 1. You will see the above procedure being applied again and again whenever you find expected values. Let's do it. First we notice that when x = 0, the entire first term vanishes, so we can rewrite the expression as: inf --- \ [L^x]*[e^(-L)] E(X) = / ---------------- --- (x-1)! x=1 Where I have made the additional simplification of cancelling the first term in the factorial with the x we had in the numerator. Now, the above looks *almost* the same as p(x), except that we have (x - 1)! instead of x!. No problem. Let us make a change-of-variable and choose y = x - 1. We'll see how everything turns out after we do that. inf if y = x - 1, then x = y + 1 --- | \ [L^(y + 1)]*[e^(-L)] E(X) = / ----------------------- --- y! y=0 | When x = 1, y = 0 Almost there - except that now we have an extra L that multiplies every term in the series. So, all we have to do is factor it out. inf --- \ [L (y)]*[e^(-L)] E(X) = L* / ------------------ --- y! y=0 And there you have it. The resulting summation is nothing more than adding up all the values of the density function of the Poisson, and we know that by the definition of the density function, it must all add up to 1. This gives us the final very neat result: E(X) = L The variance is found by very similar means, except that this time we employ another interesting little procedure (I would call it a trick, but since it is also used for finding the variance of a Binomial distribution, it's a procedure). Recall that Var(X) = E(X^2) - [E(X)]^2 Since we already know E(X), all we have to do is now find E(X^2) in order to obtain the variance. Unfortunately, this is a very difficult task, so we instead do something else. Notice that E[X(X-1)] = E(X^2 - X) = E(X^2) - E(X). This alternate expression is easier to add up, because of that factorial in the denominator of the Poisson density. Finding it will give you an expression containing E(X^2) which can then be used for finding the variance. In other symbols, E[X(X - 1] + E(X) - [E(X)]^2 = E(X^2) - E(X) + E(X) - [E(X)]^2 = E(X^2) - [E(X)]^2 = Var(X) So the only ingredient missing is E(X(X - 1)). This I'll let you find yourself. You need to add up the following infinite sum: inf --- \ [L^x]*[e^(-L)] E[X(X - 1)] = / x(x - 1) * --------------- --- x! x=0 Do it the same way I did it for the E(X). Play around with the expressions, do a substitution, and try to arrive at an expression of the form of (Something)*[Sum of p(x)]. I hope this gives you no major difficulties, once you have fully understood how to do it for E(X). I can think of an entirely different standard method for finding the mean and variance of the Poisson distribution. It involves a theoretical device called moment generating functions, abbreviated mgf's. Since this method is very straightforward, provided that we have already derived the mgf of the Poisson, I'll explain it too. The mgf of ANY distribution is given by the formula: mgf(t) = E[e^(tX)] where E( ) denotes the expected value function related to that distribution and x is a random variable. In particular, the mgf of the Poisson can be found by evaluating the above expected value (which involves a lot of fun and summing up again an infinite series). In the end, one can arrive at the following formula: mgf(t) = exp[(e^t - 1)L] where exp[ ] is another way of writing e^( ) and L is again the parameter related to the Poisson distribution, same L as before. Now, the beauty of the mgf is that it greatly simplifies calculations for obtaining means and variances. It is called _moment_ generating function, because in a sense, it packs of all the moments of the distribution into one neat expression. Moments of a random variable are E(X), E(X^2), E(X^3), etc. Notice how the mean is the first moment, E(X), and the variance is the second moment minus the first moment squared, i.e. Var(X) = E(X^2) - [E(X)]^2 There is a standard result that says that if we find the kth derivative of the mgf and evaluate it at t = 0, we will obtain the kth moment. This theorem can be obtained by expressing the mgf as an infinite sum (use the definition), then differentiating term-by-term, and finally evaluating at t = 0. Using this little theorem and the mgf of the Poisson we find that: d mgf(t) | d exp[(e^t - 1)L] | | -------- | = ----------------- | = exp[(e^t - 1)L]*Le^t | = L dt |t=0 dt |t=0 |t=0 It is reassuring to see that this agrees with our previous result. Now we need to find the second derivative in order to get the second moment. We can then use the second moment to find the variance. d^2 mgf(t) | d exp[(e^t - 1)L]*Le^t | ------------ | = ------------------------ | dt^2 |t=0 dt |t=0 | = exp[(e^t - 1)L] * (Le^t)^2 + exp[(e^t - 1)L] * Le^t | |t=0 = L^2 + L And just for an added touch of suspense, I'll let you figure out from here what the variance of a Poisson random variable is. I hope you found this explanation useful and easy to understand. If you still need more clarification, or if you have other questions of any kind, please write back. - Doctor Jordi, The Math Forum http://mathforum.org/dr.math/ |
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