# Verifying mgf for expectation, variance, and correlation

I have the following joint MGF of A and B. t1 corresponds to A and t2 corresponds to B.

I tried to find the expectation, variance, and correlation for A and B. I was wondering if the following values were correct:

E(A) = 1 Var (A) = 1

E(B) = 2 Var(B) = 2

Corr (A,B) = -1

If anyone could verify these values with me, then I would really appreciate it.

Thanks!

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## 1 Answer

In the answer to the prior question of regarding this same distribution, I gave an interpretation of $(A,B)$ as $(Z_1, Z_1+Z_2)$, where $Z_1$ and $Z_2$ were independent Poisson variable with unit mean.

Using this $\mathbb{E}(A) =\mathbb{E}(Z_1) = 1$, $\mathrm{Var}(A) =\mathrm{Var}(Z_1) = 1$. Similarly $\mathbb{E}(B) = \mathbb{E}(Z_1) + \mathbb{E}(Z_2) = 2$, and $\mathrm{Var}(B) =\mathrm{Var}(Z_1) + \mathrm{Var}(Z_2) = 2$.

Covariance $\mathrm{Cov}(A, B) = \mathrm{Cov}(Z_1, Z_1+Z_2) = \mathrm{Cov}(Z_1, Z_1) + \mathrm{Cov}(Z_1,Z_2) = \mathrm{Var}(Z_1) + 0 = 1$.

Correlation follows $\mathrm{Cor}(A, B) = \frac{1}{\sqrt{2}}$.

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Thanks! If Cor (A,B) = Cov(A,B) / sigma (A) * sigma (B), then how do you get 1? The denominator must be root 2 and I don't believe the covariance is also root 2... –  icobes Sep 29 '11 at 5:18
@icobes Yes, my bad. I have corrected the post. I was meaning to write covariance, not correlation. –  Sasha Sep 29 '11 at 5:37