Thread: Quick Quesiton with expectation and variance with independent random variables

Suppose X and Y are independent random variables with E(X) = 1 and Var(X) = 5 and E(Y) = 7 and Var(Y) = 3. Find E((2+X)^2) and Var(X-3Y-4)

For the Variance one, it's just Var(X) - 3Var(Y) - 4 right?

The quick question I had was does the square of (2+X) matter at all? Or do I just do (2 + E(X))^2?

Originally Posted by Janu42

For the Variance one, it's just Var(X) - 3Var(Y) - 4 right?

Does translation by a constant change the variance? I would encourage you to use the identity to compute it.

The quick question I had was does the square of (2+X) matter at all? Or do I just do (2 + E(X))^2?

Just expand it out and see for yourself.


Now apply the linearity and use the above formula for the variance to find .

Originally Posted by Focus

Does translation by a constant change the variance? I would encourage you to use the identity to compute it.



Just expand it out and see for yourself.


Now apply the linearity and use the above formula for the variance to find .

I know the formula for Variance, but what do you mean use the identity to compute it? I thought Var(X+Y) = Var(X) + Var(Y) for independent random variables.

Let me see if I understand. E(X) = 1 and Var(X) = 5, so 5 = E(X^2) - 1, so E(X^2) is 6. But then how does that matter to the first part? Like I said, I thought I could just plug in the Var(X) for X and Var(Y) for Y.

And for the second one, then it's 4 + 4E(X) + E(X^2)?

Originally Posted by Janu42

I know the formula for Variance, but what do you mean use the identity to compute it? I thought Var(X+Y) = Var(X) + Var(Y) for independent random variables.

Let me see if I understand. E(X) = 1 and Var(X) = 5, so 5 = E(X^2) - 1, so E(X^2) is 6. [snip] Mr F says: Yes.

And for the second one, then it's 4 + 4E(X) + E(X^2)?

Yes, and you know E(X^2).

Originally Posted by Janu42

Suppose X and Y are independent random variables with E(X) = 1 and Var(X) = 5 and E(Y) = 7 and Var(Y) = 3. Find E((2+X)^2) and Var(X-3Y-4)

For the Variance one, it's just Var(X) - 3Var(Y) - 4 right?

[snip]

Wrong. Because Var(aX + bY + c) = a^2 Var(X) + b^2 Var(Y). I can't be bothered searching but the member zorro asked the same question a few days ago - I gave a more detailed reply at that thread.