Probability Theory Proof on Expected Values that I'm having trouble with.

So this is one of the questions in my probability theory textbook and I can't figure it out, can anyone help?

Let X and Y be random variables.
a. Assuming that all of the expected values actually exist, prove that
(E(XY))2 ≤ E(X2)E(Y2)
It gave us a hint saying to consider E((aX + bY)²) and E((aX - bY)²) for arbitrary a and b, and to apply what we find to well-chosen a and b.

If you expand that out you get E(a²X² + 2abXY + b²Y²)
Or , a²E(X²) + 2abE(XY) + b²E(Y²)
but I don't know what to do with that. I realize it's quadratic in both a and b so I could try to do something with the discriminant which would get me the terms I need but I don't know if that's ok to do or how to prove it if it is.

b. If the standard deviations of X and Y are Sx and Sy, prove that |Cov(X,Y)| ≤ SxSy

I really don't know what to do for this part. I was trying to work the right hand side using the fact that the standard deviation is the root of the variance and then using the definition of variance.

Thanks in advance!

wombuster said:

So this is one of the questions in my probability theory textbook and I can't figure it out, can anyone help?

Let X and Y be random variables.
a. Assuming that all of the expected values actually exist, prove that
(E(XY))2 ≤ E(X2)E(Y2)
It gave us a hint saying to consider E((aX + bY)²) and E((aX - bY)²) for arbitrary a and b, and to apply what we find to well-chosen a and b.

If you expand that out you get E(a²X² + 2abXY + b²Y²)
Or , a²E(X²) + 2abE(XY) + b²E(Y²)
but I don't know what to do with that. I realize it's quadratic in both a and b so I could try to do something with the discriminant which would get me the terms I need but I don't know if that's ok to do or how to prove it if it is.

Click to expand...

You're on the right track here. Let's define a function
Now, like you said, this is a quadratic (in \(\displaystyle a \)), and note that it is always \(\displaystyle \geq 0 \). As you alluded to, we can use the discriminant here. In this case the discriminant \(\displaystyle 4E(XY)^2 - 4E(X^2)E(Y^2) \) must be less than or equal to zero because this quadratic can have only one real root (if any). Now you just need to do a tiny bit of algebra to get to your desired result.

Hope that helps!

wombuster said:

b. If the standard deviations of X and Y are Sx and Sy, prove that |Cov(X,Y)| ≤ SxSy

I really don't know what to do for this part. I was trying to work the right hand side using the fact that the standard deviation is the root of the variance and then using the definition of variance.

Thanks in advance!

Click to expand...

Here's a hint on this one:

One way to express the covariance of X and Y is \(\displaystyle Cov(X,Y)= E(XY) - E(X)E(Y)\). Can you see how this now relates to the inequality in your first question?