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)?

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- January 16th 2009, 09:29 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)? - January 16th 2009, 09:51 AMLast_Singularity
It is difficult to calculate variance using .

Instead, I recommend that you try this:

Of course, if are independent, the covariance term goes to zero.

So for independent variables ,

Since are given as respectively, simply recall the variance expressions for a binomial and Poisson distribution.