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I have that (X, Y) are uniformly distributed in the triangle defined by the vertices $(0, 0)$, $(1, 0)$, and $(0, 1)$. I am trying to find the following expectation value:

$$ E\left( (X-Y)^2 | X \right) $$

I am able to get the joint density, marginal densities, and conditional densities fairly easily. However, I am a little confused on how to go about solving this?

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  • $\begingroup$ As an alternative, use the second approach from here. If $(X, Y) = (U, (1 - U)V)$, then $$f_{U, V}(u, v) = 2 (1 - u) [0 < u < 1 \land 0 < v < 1], \\ \mathbb E(Y^p \mid X) = (1 - U)^p \hspace {1.5px} \mathbb E(V^p) = \frac {(1 - X)^p} {p + 1}.$$ $\endgroup$
    – Maxim
    Apr 17, 2020 at 0:16

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$$ \mathbb E\left( (X-Y)^2 | X \right)= \mathbb E\left( X^2+Y^2 -2XY | X \right)=X^2+E(Y^2|X)-2XE(Y|X)$$

Since $E(Yf(X)|X)=f(X)E(Y|X)$ Basic_properties.

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    $\begingroup$ For $E(Y|X)$, since $0 \leq y \leq 1 - x$, does the integrating remain $\int_{0}^{1-x}{y^2 f_{Y|X}(y|x) dy}$? Or must the bounds be changed? I know that if we let $Z$ be a random variable such that $Z = Y^2$, then the integration would be over $0 \leq Z \leq (1-x)^2$ but the density would be different. Would you be able to clarify this? $\endgroup$
    – Eoin S
    Apr 15, 2020 at 23:18
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    $\begingroup$ $E(Y^k|X=x)=\int_0^{1-x} y^k f(y|x) dy$. You should consider that $f(y|x)$ calculated by $f(x,y)$ so $y<1-x$. $\endgroup$
    – Masoud
    Apr 16, 2020 at 0:56

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