If $Y|X=x\sim N(x,x^2)$ and $X\sim U(0,1)$ is it true that $EX=E(Y|X)$?

Question

I tried to apply the definition of conditional expectation to it but with no success. Any hint, anyone?

Answer 1

Hint

We have $E(Y\mid X=x)=x$ since $Y$ is conditionally $N(x,x^2).$ Thus $E(Y\mid X) = X.$ So... does $E(Y\mid X) = E(X)?$

Answer 2

Let $X$ and $Y$ be jointly continuous random variables for which expectations $E(X)$ and $E(Y)$ exist. Then we have the following definitions: $$g(x) = E(Y|x) = \int_{-\infty}^\infty yf_{Y|x}(y|x)\,dy.$$ Then $E(Y|X) = g(X),$ so that $E(Y|X)$ is a function of the random variable $X,$ while $E(X)$ is a number.

Furthermore, it is a theorem that $E[E(Y|X)] = E(Y).$ Please look for that in your textbook or class notes.

For your particular joint distribution, $g(x) = x,$ for $0 < x < 1.$ [See the answer by @spaceisdarkgreen (+1)]. The following simulation in R provides a figure that roughly illustrates the joint distribution and the expectations. [I assume you mean $SD(Y|X=x) =x.$]

m = 60000;  x = y = runif(m)
for(i in 1:m) {
  y[i] = rnorm(1, x[i], x[i]) }
plot(x, y, pch=".", xaxs="i", yaxs="i")
  abline(0, 1, col="green", lwd=3)
  abline(h=1/2, col="red"); abline(v=1/2, col="red")
mean(x);  mean(y)
## 0.5001309  # aprx E(X) = 1/2
## 0.4959868  # aprx E(Y) = 1/2