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Let $X_1 , X_2$ be iid random variables, following the pdf: $$ f_\theta (x) = \theta x^{\theta-1} $$ for $\theta >0$ and $0 <x<1$. Let $$T=X_1 X_2$$ $$ Y = \begin{cases} 1, & \text{if } X_1 > \dfrac{1}{2} \\ 0, & \text{otherwise} \end{cases} $$ I need to find conditional density $Y|T$

My work is: The conditional density of $Y$ given $T$ is given by the formula $$f_{X|T}(x_1|t)=\dfrac{f_{Y,T}(x_1,t)}{f_T(t)}$$

The joint distribution is Let $$x_1=x_1 \text{ and } T=x_1 x_2 \text{ Jacobian is } J= \frac{1}{x_1}$$

$$f_{Y,T}(x_1,t)= \frac{1}{x_1} f_{Y}(x_1) f_{T}(\frac{t}{x_1}) = \frac{1}{x_1} \theta (x_1)^{\theta-1} * \theta (\frac{t} {x_1})^{\theta-1}= \frac{1}{x_1} \theta^2 t^{\theta-1} $$

and the marginal density of $T$ is

$$f_T(t)=\int_\frac{1}{2}^1\ \frac{1}{x_1}\theta^2 t^{\theta-1} \text{d}x_1 = \theta^2 t^{\theta-1} ln(2) $$

Now, the conditional density $ = \dfrac{1}{x_1} \ln(2)$

I'm integrated the result, and I get $1$, so it's pdf But I'm confused because I think the conditional density must be function of $t$.

Thanks alot.

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  • $\begingroup$ First remark: the pdf of $T$ is $\theta^2 t^{\theta-1}\ln(1/t)$. It is a pdf of $e^{-Y}$ where $Y$ is Gamma distributed with $\alpha=2$, $\beta=\theta$. $\endgroup$ – NCh Mar 31 at 14:04
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$Z=-\ln X\sim Exponential (\theta)$ ($f_Z(z)=\theta e^{\theta Z}$)

$E(Y|T)=P(X>\frac{1}{2}|T)$

$P(X_1>\frac{1}{2}|X_1 X_2=t)=P(-\ln X_1 <\log(2)|-\ln X_1 -\ln X_2=-\ln t)$

$=P(Z_1 <\ln 2|Z_1 +Z_2=-\ln t)$

this is UMVUE for $P(Z_1 <\ln 2)$ since $Z_1 +Z_2$ is sufficient and complete estimator.let $w=-\ln t$ (2)

$P(Z_1 <\log(2)|Z_1 +Z_2=w)=P(\frac{Z_1}{Z_1 +Z_2} <\frac{\ln 2}{w} |Z_1 +Z_2=w)$

note $\frac{Z_1}{Z_1 +Z_2}\sim Beta(1,1)=U(0,1)$ and does not depend on $\theta$ so by Basu theorem is independent of $Z_1 +Z_2$ so

$=P(\frac{Z_1}{Z_1 +Z_2} <\frac{\ln 2}{w})=P(U(0,1)< \frac{\ln 2}{w})=\frac{\ln 2}{-\ln t}=\frac{\ln 2}{\ln \frac{1}{t}}$

note $0\leq \frac{\ln 2}{-\ln t} \leq 1$ and $0 \leq t \leq 1$

so $\frac{\ln 2}{-\ln t} \leq 1$ implies $t\geq \frac{1}{2}$

by using (2) we can utilize another approach to find $E(Y|T)$

your method

in the marginal density , you can not integrate like $$f_T(t)=\int_\frac{1}{2}^1\ \frac{1}{x_1}\theta^2 t^{\theta-1} \text{d}x_1$$

this is a plot of $XY=t$ for varoius $t$ so $\int_\frac{1}{2}^1$ is not logic

<span class=$$XY=t$$">

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