# Show T is sufficient for exponential statistic using conditional distribution approach.

Given iid $X_1,\dots X_n \sim \exp(\beta)$, use the conditional distribution approach to show that $T=\sum_{i=1}^n X_i$ is sufficient for $\beta$.

My attempt: $\mathbb{P}\left(\bigcap_{i=1}^n (X_i = x_i)\mid T=t\right) = 0$ if $\sum_{i=1}^nx_i\ne t$, but then we have that the event that $\bigcap_{i=1}^n (X_i = x_i)$ is a subset of the event that $T=\sum_{i=1}^n X_i=t$. So

\begin{align}\mathbb{P}\left(\bigcap_{i=1}^n (X_i = x_i)\mid T=t\right) &= \frac{\mathbb{P}\left(\bigcap_{i=1}^n (X_i = x_i) \cap T=t\right)}{\mathbb{P}(T=t)} =\frac{\mathbb{P}\left(\bigcap_{i=1}^n (X_i = x_i)\right)}{\mathbb{P}(T=t)}\end{align} with \begin{align}\mathbb{P}\left(\bigcap_{i=1}^n (X_i = x_i)\right)&=\prod_{i=1}^n \beta e^{-x_i\beta}\\[0.2cm]\mathbb{P}(T=t) &= \frac{\beta^n x^{n-1}e^{-n\beta}}{\Gamma(n)}\end{align}

Now the problem is, I don't see how to get rid of the $\beta$.

EDIT:

$$\mathbb{P}\left(\bigcap_{i=1}^n (X_i = x_i)\mid T=t\right) = \frac{e^{n\beta}\prod_{i=1}^n e^{-x_i\beta}}{\frac{ x^{n-1}}{\Gamma(n)}}$$

Still got stuck getting rid of the exponential $\beta$

• The pdf of $\exp(\beta)$ is $f(x\mid \beta)=\beta e^{-x\beta}$ and not with $\frac1β$ as you have it. Right? – Jimmy R. Oct 31 '16 at 14:45
• @JimmyR. You are right! Sorry for such a stupid mistake. I can not see the solution. – Dean332 Oct 31 '16 at 14:48
• Do you know the factorization theorem? – Jimmy R. Oct 31 '16 at 14:50
• @JimmyR. This is just an exercise to be done before the factorisation theorem – Dean332 Oct 31 '16 at 14:53

Your mistake is in $P(T=t)$. You have $e^{-nβ}$ in the numerator, but it should be $e^{-t\beta}$. Specifically,
$$P(T=t)=\frac{\beta^nt^{n-1}e^{-\beta t}}{\Gamma(n)}$$
Also, observe that $$\prod_{i=1}^ne^{-x_i\beta}=e^{-x_1\beta}e^{-x_2\beta}\dots e^{-x_n\beta}=e^{-\beta \sum_{i=1}^nx_i}=e^{-\beta T(X)}$$