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The random variable X has a poisson distribution with mean 5. The random variable Y is...

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leevicboy | eNoter

Posted August 31, 2013 at 11:40 AM via web

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The random variable X has a poisson distribution with mean 5. The random variable Y is defined as Y=2^X. Find E(Y).

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mathsworkmusic | (Level 3) Associate Educator

Posted October 9, 2013 at 8:55 PM (Answer #1)

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We have that the random variable `X` is such that

`X ~`~ Pois(`lambda`)  where the mean is given by `lambda=5` .

We would like to know the expected value of the random variable

`Y = 2^X`

Using the fact that the expectation of a function of `x`, `g(x)`, can be calculated as

`E[g(X)] = Sigma_x g(x) Pr[X=x]`

we then have that

`E[Y] = E[2^X] = Sigma_x (2^x) (lambda^x e^(-lambda))/(x!)`

Rearranging by gathering terms to the power of `x` we have that

`E[Y] = Sigma_x((2lambda)^xe^(-lambda))/(x!)`

The next step is to rearrange the expression within the sum so that it is made up of a Poisson distribution and a constant term (not involving `x` ). In the first element under the sum the `lambda` has now become `2lambda` , so we carry that through to the next element, giving

`E[Y] = Sigma_x((2lambda)^xe^(-2lambda))/(x!) (e^(-lambda)/e^(-2lambda))`

The constant term at the end can now come to the front of the summation as it doesn't involve `x`. The sum of the remaining term in the summation is 1, since the term is a Poisson distribution (specifically with mean `2lambda`). 

Therefore we finally have that

`E[Y] = (e^(-lambda)/e^(-2lambda)) = e^(2lambda-lambda) = e^lambda`

Here, `lambda=5` so that `E[Y] = E[2^X] = e^5 = 148.4132`

Answer

E[Y] = exp(5) = 148.4132

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