# Find conditional distribution of Y|T

Let $$X_1 , X_2$$ be iid random variables, following the pdf: $$f_\theta (x) = \theta x^{\theta-1}$$ for $$\theta >0$$ and $$0 . Let $$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.

• 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 $\alpha=2$, $\beta=\theta$. – NCh Mar 31 at 14:04

## 1 Answer

$$Z=-\ln X\sim Exponential (\theta)$$ ($$f_Z(z)=\theta e^{\theta Z}$$)

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

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

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

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 $$XY=t$$">