Conditional distribution of $X_1\mid -\sum\log X_i$ where $X_i$'s are i.i.d with pdf $\theta x^{\theta-1}$

Question

$X_1, X_2, ..., X_n$ are i.i.d. with p.d.f. $$\theta x^{\theta-1}I_{\{0,1\}}(x)$$ Let $S = -\sum_{i=1}^{n}\log X_i.$ I want to find the conditional distribution $f_{X_1|S=s}(x_1\mid s)$.

I know that $-\log X_i$ has the exponential distribution with parameter $\theta$ and $S$ has Gamma$(n,\theta)$ distribution.

This is what I have done:

$\Delta x_1 (f_{X_1|S=s}(x_1|s)) = \Delta x_1\frac{f_{X_1,S}(x_1,s)\Delta s}{f_S(s)\Delta s} \approx \frac{P(x_1 < X_1 < x_1 + \Delta x_1 ; s < S < s + \Delta s)}{P(s < S < s + \Delta s)} = \frac{P(-\log(x_1 + \Delta x_1) < -\log X_1 < -\log(x_1) + \Delta x_1 ; s < -\sum_{i=1}^{n}\log X_i < s + \Delta s)}{P(s < S < s + \Delta s)} = \frac{P(-\log(x_1 + \Delta x_1) < -\log X_1 < -\log(x_1) + \Delta x_1 ; s + \log(x_i) < -\sum_{i=2}^{n}\log X_i < s + \log(x_i) + \Delta s)}{P(s < S < s + \Delta s)} = \frac{P(-\log(x_1 + \Delta x_1) < -\log X_1 < -\log(x_1) + \Delta x_1)P(s + \log(x_i) < -\sum_{i=2}^{n}\log X_i < s + \log(x_i) + \Delta s)}{P(s < S < s + \Delta s)} = \frac{(\Delta x_1 \theta e^{-\theta \log(\frac{1}{x_1})})(\Delta s \frac{\theta ^ {n-1}}{\Gamma (n-1)}(s + \log(x_1))^{n-2}e^{-\theta (s+ \log(x_1))}) }{\Delta s\frac{\theta^n}{\Gamma(n)}s^{n-1}e^{-\theta s} } = \Delta x_1\frac {(n-1)(s+ \log x_1)^{(n-2)}}{s^{n-1}}$

This is very nearly correct, I know that the correct answer should be

$$\Delta x_1\frac {(n-1)(s+ \log x_1)^{(n-2)}}{x_1s^{n-1}}$$

Can anyone spot my mistake?

Answer 1

Although you are proving from first principles (+1 for that), you can also first find the distribution of $-\ln X_1\mid S$, and hence find that of $X_1\mid S$ using the familiar transformation method.

Define $$U=-\ln X_1\quad,\quad V=-\sum_{i=2}^n \ln X_i$$

Then we are looking for the conditional distribution $$(-\ln X_1\mid S)=U\mid U+V $$

Clearly, $U$ and $V$ are independently distributed, and it is easy to verify that $U$ has an Exponential distribution with mean $1/\theta$ and $V$ is a $\text{Gamma}(n-1,\theta)$ variable with density

$$f_V(v)=\frac{\theta^{n-1}}{\Gamma(n-1)}e^{-\theta v}v^{n-2}\mathbf1_{v>0}$$

And since $U$ (also a Gamma variable) and $V$ are independent, $U+V\sim \text{Gamma}(n,\theta)$.

Using a simple change of variables, the joint distribution of $(U,U+V)$ is

$$f_{U,U+V}(u,s)=\frac{\theta^n}{\Gamma(n-1)}e^{-\theta z}(s-u)^{n-2}\mathbf1_{0<u<s}$$

This implies that the conditional density of $U\mid U+V=s$ is

$$f_{U\mid U+V}(u\mid s)=\frac{(n-1)(s-u)^{n-2}}{s^{n-1}}\mathbf1_{0<u<s}$$

A final change of variables yields the density of $(e^{-U}\mid U+V=s)=(X_1\mid S=s)$ :

\begin{align} f_{X_1\mid S}(x_1\mid s)&=f_{U\mid U+V}(-\ln x_1\mid s)\left|\frac{du}{dx_1}\right|\mathbf1_{\exp(-s)<x_1<1} \\\\&=\frac{(n-1)(s+\ln x_1)^{n-2}}{x_1s^{n-1}}\mathbf1_{\exp(-s)<x_1<1} \end{align}