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:

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

Answer 2

You can use the fact that the expectation of a variate's multiple is multiple of expectation of that variate.

The values are:

$E(W) = 4np + 6mp\\Var(W) = 16np(1-p) + 36mp(1-p)$

What are the expectation and variance of a binomial variate?

Let the variate K belongs to binomial distribution with parameters n and p.

$K \sim B(n,p)$

Then its expectation and variance are given by:

$E(K) = np\\Var(K) = np(1-p)$

How to calculate the variance and expectation of variate W?

Since it is given that  $X \sim B(n,p), Y \sim B(m, p)$

thus,

$E(X) = np, Var(X) = np(1-p)\\E(Y) = mp, Var(Y) = mp(1-p)$

It is a fact that:

$E(aX) = aE(X)$ ,

$Var(aX) = a^2Var(X)$   where a is a constant.

$E(W) = E(4X + 6Y) = 4E(X) + 6E(Y) = 4 \times np + 6 \times mp \\\\Var(W) = Var(4X + 5Y) = 4^2Var(X) + 6^2 Var(Y) \\Var(W) = 16 \times np(1-p) + 36 \times mp(1-p)$

Thus,

The values are:

$E(W) = 4np + 6mp\\Var(W) = 16np(1-p) + 36mp(1-p)$

Learn more about expectation and variance here:

https://brainly.com/question/26062265