# Math Help - Conditional Expectation of 2 independent standard normal random variables

1. ## Conditional Expectation of 2 independent standard normal random variables

I came across this question on some web-site:

X and Y are 2 independent standard normal random variables. Find E(X | X + Y = 1).

I'm thinking:

E(X | X + Y = 1) = $\int x P(X | X + Y = 1)$
= $\int x \frac{P(X=x, Y=1-x)}{P(Y=1-x)}$
= $\int x \frac{P(X=x) P(Y=1-x)}{P(Y=1-x)}$ since X and Y are independent
= $\int x P(X=x)$
= E(X)

I'd like to know if this is correct? Thanks.

2. ## Re: Conditional Expectation of 2 independent standard normal random variables

If x+y=1 then x=1-y.
Expected values are additive so E(X)= E(1-Y)= E(1) - E(Y)

3. ## Re: Conditional Expectation of 2 independent standard normal random variables

Hi Shakarri,

E(X) = E(1) - E(Y) would mean E(X) = 1 - 0 = 1 (Since Y is a standard normal random variable, E(Y) = 0)

But X is also a standard normal random variable, so E(X) = 0 . So, it's still not clear to me.
Could you explain? Thanks!

4. ## Re: Conditional Expectation of 2 independent standard normal random variables

Oh I missed that you said they were both standard random variables. Well you stated in the problem that they are independent but since they sum to 1 they can't be independent. It doesn't make sense to say that they are both standard normal variables and they must sum to 1.

Perhaps the question is meant to be the following scenario: You take two standard normal random variables, if their sum is not 1 then ignore that sample, if their sum is 1 then take that sample. What is the expected value of X in the samples which are taken?
If this is the case then although X is a standard normal variable, the samples which are not ignored are not a standard normal variable.

5. ## Re: Conditional Expectation of 2 independent standard normal random variables

The question never said that X + Y = 1. (X + Y = 1 only appears in the conditional expectation).
So, how would we calculate E(X | X + Y = 1) if that was the case?

6. ## Re: Conditional Expectation of 2 independent standard normal random variables

Originally Posted by JohnDoe2013
The question never said that X + Y = 1. (X + Y = 1 only appears in the conditional expectation).
So, how would we calculate E(X | X + Y = 1) if that was the case?
The trouble that I have with this is that $Pr[X+Y=1]=0$.

The loci of points in the X,Y plane where this is true is a 1D line of zero 2D measure.

The probability of these points is the 2D integral 2D standard normal distribution along this line and this integral is 0.

So I don't see how to obtain a non-zero distribution of $p(X | X+Y=1)$

My intuition says that something conditioned on something impossible (of probability 0) can never happen and is thus also probability 0.

What might be instructive is to let $1-\epsilon <= X+Y <= 1+\epsilon$, find the marginal of X given that, now possible, condition, do the math from there and see what happens in the limit as $\epsilon \to 0$

This could all be incorrect and maybe one of the powerhouses will chime in with some wisdom.

7. ## Re: Conditional Expectation of 2 independent standard normal random variables

romsek, thanks. I'm inclined to think that there is something wrong with the question. I'm closing this thread.

8. ## Re: Conditional Expectation of 2 independent standard normal random variables

Originally Posted by romsek
The trouble that I have with this is that $Pr[X+Y=1]=0$.

The loci of points in the X,Y plane where this is true is a 1D line of zero 2D measure.

The probability of these points is the 2D integral 2D standard normal distribution along this line and this integral is 0.

So I don't see how to obtain a non-zero distribution of $p(X | X+Y=1)$

My intuition says that something conditioned on something impossible (of probability 0) can never happen and is thus also probability 0.

What might be instructive is to let $1-\epsilon <= X+Y <= 1+\epsilon$, find the marginal of X given that, now possible, condition, do the math from there and see what happens in the limit as $\epsilon \to 0$

This could all be incorrect and maybe one of the powerhouses will chime in with some wisdom.
I don't know if this is a good example, but suppose you wanted to pick random numbers from the set of all integers. Suppose X=1 if you pick the number zero and X=0 otherwise. Suppose Y=1 if you pick the number one and Y=0 otherwise. P(X+Y=1) = 0. But, $P(X|X+Y=1) = \dfrac{1}{2}$. Am I wrong? Is this a bad example?

9. ## Re: Conditional Expectation of 2 independent standard normal random variables

Originally Posted by romsek
The trouble that I have with this is that $Pr[X+Y=1]=0$.

The loci of points in the X,Y plane where this is true is a 1D line of zero 2D measure.

The probability of these points is the 2D integral 2D standard normal distribution along this line and this integral is 0.

So I don't see how to obtain a non-zero distribution of $p(X | X+Y=1)$

My intuition says that something conditioned on something impossible (of probability 0) can never happen and is thus also probability 0.

What might be instructive is to let $1-\epsilon <= X+Y <= 1+\epsilon$, find the marginal of X given that, now possible, condition, do the math from there and see what happens in the limit as $\epsilon \to 0$

This could all be incorrect and maybe one of the powerhouses will chime in with some wisdom.
I looked at the limit I mention and it does indeed go to zero as expected.