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I'm having a hard time getting conditional hope. I know that $$P(X = x | Y = y) = y^x(1-y)^{1-x}, \:\: x = \{0,1\}, \:\: 0 \leq y \leq 1.$$ Besides that, $Y \sim U[0, 1]$.

I want to get $E(Y|X=x).$

Starting from the definition of conditional probability, I arrive at

\begin{align} P(Y = y | X = x) = & \frac{P(Y=y, X=x)}{P(X=x)}\\ & \frac{P(X=x|Y=y)P(Y=y)}{P(X=x)} \end{align}

Adding in $Y$ I get that $X \sim Bern(0,5)$. That way,

\begin{align} P(Y = y | X = x) = & \: \frac{P(X=x|Y=y)P(Y=y)}{P(X=x)}\\ = & \: \frac{y^x(1-y)^{1-x}yI(0 \leq y \leq 1)}{(\frac{1}{2})^x(\frac{1}{2})^{1-x}}\\ = & \: 2y^{x+1}(1-y)^{x-1}I(0 \leq y \leq 1) \end{align}

Where is the misconception, since for $x = 0$, $Y|x=0$ is not a density?

Thanks in advance!

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I think you have included an additional $y$ in the numerator and the index for $(1-y)$ seems to be not correct as well.

$$f_Y(y|X=x) = \frac{y^x(1-y)^{1-x}}{\frac12}=2y^x(1-y)^{1-x}$$

Hence,

$$f_Y(y|X=0) = 2(1-y)$$ which is nonnegative and integrates to $1$.

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  • $\begingroup$ True, now that I realize that I have replaced the accumulated by density. Thank you! $\endgroup$ Feb 26 '20 at 13:29

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