show that (assuming all the expectatons and variances exist and are finite)

E(Y)= E(E[Y|X])

and

Var (Y) = E(Var[Y|X]) + Var (E[Y|X])

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- Oct 5th 2008, 03:38 PMweakmathdouble expectation proof
show that (assuming all the expectatons and variances exist and are finite)

E(Y)= E(E[Y|X])

and

Var (Y) = E(Var[Y|X]) + Var (E[Y|X]) - Oct 6th 2008, 12:12 AMmr fantastic
Let X and Y have have joint density function f(x, y) and marginal density functions $\displaystyle f_X (x)$ and $\displaystyle f_Y (y)$ respectively.

Then:

$\displaystyle E(Y) = \int_{-\infty}^{+\infty} y \, f_Y (y) \, dy = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} y \, f(x, y) \, dx \, dy$

$\displaystyle = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} y \, f(y | x) \, f_X (x) \, dy \, dx $

$\displaystyle = \int_{-\infty}^{+\infty} \left[ \int_{-\infty}^{+\infty} y \, f(y | x) \, dy \right] f_X (x) \, dx $

$\displaystyle = \int_{-\infty}^{+\infty} E(Y | X = x)\, f_X (x) \, dx $

$\displaystyle = E[E(Y | X)]$.

Var(Y) is left for you to try doing again.