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There are two random variables $X$ and $Y$, both of them distributed standard normal. Based on $X$ and $Y$, define $Z = X + a Y$ for $a$ a known parameter.

I'm interested in $E(X | Z)$.

It seems to me that a starting point could be the approach taken here. So I start with:

$ Z = E( X | Z) + a E(Y | Z) $

and then I write out the expectation terms explicitly, hoping to find an easy substitution:

$ E(X | Z) = \frac{\int_{-\infty}^{\infty}x f(x)f(\frac{Z-x}{a})dx}{\int_{-\infty}^{\infty} f(x)f(\frac{Z-x}{a})dx} $

and

$ E(Y | Z) = \frac{\int_{-\infty}^{\infty}y f(y)f(Z-ay)dy}{\int_{-\infty}^{\infty} f(y)f(Z-ay)dy} $

But here I'm stuck, I can't find an substition that would allow me to express $E(Y|Z)$ in terms of $E(X|Z)$, in order to plug in above. So how to continue from here?

Edit:

For $a =1$, one can simply use $E(X|Z) = E(Y | Z)$, plug in above, and obtain $E(X|Z) = Z/2$.

I conjecture that in general, it holds that $$E(X|Z) = \frac{Z}{1 + a^2}.$$ This is supported by several numerical simulations I made.

This would require showing that $E(Y|Z) = a E(X | Z)$. In case substituting in the integral does not help, how else could I show that?

bonifaz
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  • Nope. That is not going anywhere helpful. With a change of variables we find that:$$\begin{align}\mathsf E(Y\mid Z)&=\dfrac{\int_\Bbb R y f(y)f(Z-ay),\mathrm d y}{\int_\Bbb R f(y)f(Z-ay),\mathrm d y}\&=\dfrac{\int_\Bbb R \tfrac{Z-x}a f(\tfrac{Z-x}a)f(x),\mathrm d \tfrac{Z-x}a}{\int_\Bbb R f(\tfrac{Z-x}a)f(x),\mathrm d \tfrac{Z-x}a}\&=\frac Za-\dfrac{\int_\Bbb R xf(\tfrac{Z-x}a)f(x),\mathrm d x}{a\int_\Bbb R f(\tfrac{Z-x}a)f(x),\mathrm d x}\&=\tfrac 1a(Z-\mathsf E(X\mid Z))\[2ex]\therefore\quad Z&=\mathsf E(X\mid Z)+a\mathsf E(Y\mid Z)&\text{uh...yeah}\end{align}$$ – Graham Kemp Mar 05 '21 at 00:12
  • Yes I see, too bad. Well then maybe some other substitution? Or using symmetry of the normal PDF? – bonifaz Mar 08 '21 at 08:38

1 Answers1

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We first need to find the joint distribution of $X$ and $Z$. This can easily be found since the $cov(X,Z)=1$ and hence

$$ \left(\begin{array}{cc} X \\ Z \end{array} \right) \sim \mathcal{N}\left(\left(\begin{array}{cc} 0 \\ 0 \end{array} \right) , \left(\begin{array}{cc} 1 & 1 \\ 1 & 1+a^2 \end{array} \right) \right)$$

Now we can simply apply the formula for the conditional expectation from a bivariate normal distribution and obtain

$$E(X | Z) = \frac{Z}{1+a^2}.$$

Some background notes:

Lecture notes on multivariate normal distribution

Marginal and conditional distributions of multivariate normal distribution

and Chapter 3 (Theorem 3.3.4) in Tong, Y.L., 2012. The multivariate normal distribution. Springer Science & Business Media.

bonifaz
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