Question

Let X be a binomial random variable based on n trials and a success probability of pX; let Y be an independent binomial random variable based on m trials and a success probability of pY. Find E(W) and Var(W), where W=4X+6Y.

Answer 1

Answer:

$E(W) = 4 [np_x] + 6 [m p_y]$

$Var(W) = 16[np_x (1-p_x)] +36 [mp_y (1-p_y)]$

Step-by-step explanation:

For this case we have the following distributions given:

$X \sim Binom (n, p_x)$

The expected value for x is $E(X) = n p_x$

And the variance $Var(X) = np_x (1-p_x)$

$Y \sim Binom (m, p_y)$

The expected value for x is $E(Y) = n p_y$

And the variance $Var(Y) = mp_y (1-p_y)$

We define a new random variable:

$W = 4X +6Y$

We need to find the expectd value and the variance for W.

For the expected value we have this:

$E(W) = E(4X+6Y) = E(4X) + E(6Y) = 4E(X) + 6E(Y)$

And if we replace the parameters we got:

$E(W) = 4 [np_x] + 6 [m p_y]$

Now foe tha variance of W we know that X and Y are indpenendet variable so then $Cov(X,Y) =0$

$Var(W)= Var(4X +6Y) = Var(4X) +Var(6Y) + 2 Cov(X,Y)$

For this case we can use the property that is a is a constant and X a random variable then $Var(aX) = a^2 Var(X)$, and we got this:

$Var(W) = 16 Var(X) + 36 Var(Y)$

And if we replace we got:

$Var(W) = 16[np_x (1-p_x)] +36 [mp_y (1-p_y)]$