Formula for conditional variance for continuous random variable.

Question

I need to expand the conditional variance of continuous random variable as a sum of integrals. Here is my try:

$$D(Y|X)=E[Y^2|X] - [E(Y|X)]^2 + EY = \int_y y^2f(x|y)dy -\left(\int_y yf(x|y)\right)^2dy $$ But I know that the second term is incorrect / could be rewritten otherwise. Maybe somoene has an insight?

Where did you get this equation? Why $\mu_y$ term? In any case the second term should not have $y^2$, only $y$. — herb steinberg, Feb 13 '21 at 23:47
@herbsteinberg Ok, now I corrected the equation. Maybe you know how to rewrite / expand the second term? — user, Feb 14 '21 at 10:17
tommik answer is correct. You need to move $dy$ inside the square of the integral. — herb steinberg, Feb 14 '21 at 16:38

score 1 · Accepted Answer · answered Feb 14 '21 at 10:45

Your formula of conditional variance is wrong...the correct one is the following

$$\mathbb{V}[Y|X=x]=\mathbb{E}[Y^2|X=x]-\mathbb{E}^2[Y|X=x]=$$

$$=\int_{-\infty}^{+\infty}y^2f_{Y|X}(y|x)dy-\left[\int_{-\infty}^{+\infty}yf_{Y|X}(y|x)dy \right]^2$$

There is nothing else to expand...

Formula for conditional variance for continuous random variable.

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