Conditional expectation application

Question

Suppose I have two non-negative r.v.s $X,Y \geq 1$ and I know $\mathbb{E}(X^2 | Y) = (Y-1) ^ 2$.

Does this mean $\mathbb{E}(X | Y) = Y - 1$?

Cause, if i recall right the intro to probability, I have $\mathbb{E}(X^2 | Y = y) = (y - 1)^2$, so it obvious that $\mathbb{E}(X | Y = y) = y - 1$ for all $y$. But then I saw that conditional expectation is actually a function - projection to some space etc. and now this doesn't seem obvious at all.

Answer 1

No.

Let it be that $Z$ is a random variable with $Z>0.5$, $\mathbb EZ^2=1$ and $0<\mathbb EZ<1$. Further let it be that $Y$ and $Z$ are independent and $Y>3$.

Then for $X=(Y-1)Z>1$ and we find: $$\mathbb E[X^2\mid Y]=\mathbb E[(Y-1)^2Z^2\mid Y]=(Y-1)^2\mathbb EZ^2=(Y-1)^2$$ and:$$\mathbb E[X\mid Y]=\mathbb E[(Y-1)Z\mid Y]=(Y-1)\mathbb EZ<(Y-1)$$

Answer 2

Generally this is not true. Even without introducing any conditioning, in all generality $\mathbf{E}[X^2] \neq (\mathbf{E}[X])^2$.

You can still say something: from Jensen's inequality, since $x \mapsto x^2$ is convex, you always have that $\mathbf{E}[X^2] \geq (\mathbf{E}[X])^2$, which also applies to your original question for conditional probabilities:

$$\mathbf{E}[X | Y] \leq \sqrt{\mathbf{E}[X^2 | Y]} = |Y-1| = Y - 1 \text{ from } Y \geq 1$$

Answer 3

Let $\Omega=\{a,b\}, \mathcal{F}=2^\Omega$ and $\mathbb{P}$ the uniform probability on $\Omega$. Define $X(a)=\sqrt{3/2}, X(b)=\sqrt{5/2}$ and $Y(a)=\sqrt2+1=Y(b)$. Then $X,Y\ge1$ and $$\mathbb{E}(X^2|Y)=\mathbb{E}(X^2)=\frac{1}{2}\frac{3}{2}+\frac{1}{2}\frac{5}{2}=2=(Y-1)^2$$ but

$$\mathbb{E}(X|Y)=\mathbb{E}(X)=\frac{1}{2}\sqrt{\frac{3}{2}}+\frac{1}{2}\sqrt{\frac{5}{2}}\neq\sqrt{2}=Y-1$$

Answer 4

Let $X_0$ be $N(0,1)$, $X=X_0^{2}+1$ and $Y=1+\sqrt {3+E(X_0)^{4}}$. Then $E(X^{2}|Y)=(Y-1)^{2}$ and $E(X|Y)=2\neq Y-1$ My idea is to give a simple example where $Y$ is a constant.