# expected value

• Jan 16th 2009, 08:29 AM
lord12
expected 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 AM
Last_Singularity
Quote:

Originally Posted by lord12
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)?

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.