Hi, so the definition of $\displaystyle E(g(Y_1)|Y_2=y_2) = \int_{-\infty}^{\infty} g(y_1) f(y_1|y_2) dy_1$ however how do we define $\displaystyle E(g(Y_1)|Y_2 \le y_2)$ ie, when the condition is an inequality rather than an equality?

Thanks

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- Feb 13th 2013, 07:39 PMusagi_killerConditional expectation with inequality
Hi, so the definition of $\displaystyle E(g(Y_1)|Y_2=y_2) = \int_{-\infty}^{\infty} g(y_1) f(y_1|y_2) dy_1$ however how do we define $\displaystyle E(g(Y_1)|Y_2 \le y_2)$ ie, when the condition is an inequality rather than an equality?

Thanks - Feb 14th 2013, 03:19 PMchiroRe: Conditional expectation with inequality
Hey usagi_killer.

For the inequality condition what is done is we find a random variable that fits the definition of Y2 <= y2.

The easiest way to do this is have a double integral with the limits of Y2 going from -infinity to y2.

When you evaluate this double integral you will get an expression in terms of y2. - Feb 14th 2013, 06:18 PMusagi_killerRe: Conditional expectation with inequality
Thanks chiro, so what would the expression look like? Do we still use the conditional density function?

Would it be:

$\displaystyle E(g(Y_1)|Y_2 \le y_2) = \int_{-\infty}^{y_2} \int_{-\infty}^{\infty} g(y_1) f(y_1|y_2) dy_1 dy_2$

thanks - Feb 14th 2013, 08:12 PMchiroRe: Conditional expectation with inequality
Yes that looks right.