Suppose X1 ~ Binomial(n = 15; p = 1=3) and X2 ~ Poisson(lambda = 4) and X1 and X2 are independent. Let Y = 3X1 + 4X2.

Then E[Y] = 3E[X1] + 4E[X2]. But Var(Y) = E(Y^2)- E(Y)^2. How do I find E(Y^2)?

Printable View

- Jan 16th 2009, 08:29 AMlord12expected value
Suppose X1 ~ Binomial(n = 15; p = 1=3) and X2 ~ Poisson(lambda = 4) and X1 and X2 are independent. Let Y = 3X1 + 4X2.

Then E[Y] = 3E[X1] + 4E[X2]. But Var(Y) = E(Y^2)- E(Y)^2. How do I find E(Y^2)? - Jan 16th 2009, 08:51 AMLast_Singularity
It is difficult to calculate variance using $\displaystyle E[Y^2]$.

Instead, I recommend that you try this:

$\displaystyle var(aX_1 + bX_2) = a^2 var(X_1) + b^2 var(X_2) + 2 ab cov(X_1,X_2)$

Of course, if $\displaystyle X_1,X_2$ are independent, the covariance term goes to zero.

So for independent variables $\displaystyle X_1,X_2$,

$\displaystyle var(aX_1 + bX_2) = a^2 var(X_1) + b^2 var(X_2)$

Since $\displaystyle a,b$ are given as $\displaystyle 3,4$ respectively, simply recall the variance expressions for a binomial and Poisson distribution.