# Proving that $T(X)$ is a sufficient statistic for $\theta$

I have that $$X=(X_1,...,X_n)$$ is a rv consisting of $$n$$ iid exponential rv's where $$\theta$$ is the parameter (and thus mean $$1\over \theta$$) .

I have to prove that $$T(X)=\sum^n_{i=1}X_i$$ is a sufficient statistic for $$\theta$$ using the theorem in my textbook which states that if $$\frac{f_{\theta}(x)}{f^T_{\theta}(T(x))}$$ is constant in $$\theta$$ then $$T(X)$$ is a sufficient statistic for $${\theta}$$.

So what I've done is:

First I calculated $$f_{\theta}(x)= \theta^n e^{-\theta\sum^n_{i=1}X_i}$$ since $$x=(x_1,...,x_n)$$ are iid.

Then I calculated $$f_{\theta}^T(T(x))= \frac{{\sum^n_{i=1}X_i}^{n-1}e^{\frac{\sum^n_{i=1}}{\theta}}}{\theta^n(n-1)!}$$ since $$T(X)\sim gamma(n,\theta)$$

So now using the theorem with the ratio I get stuck because the $$\theta$$'s don't cancel out:

$$\frac{f_{\theta}(x)}{f^T_{\theta}(T(x))}=\frac{\theta^n e^{-\theta\sum^n_{i=1}X_i}}{\frac{{\sum^n_{i=1}X_i}^{n-1}e^{\frac{\sum^n_{i=1}}{\theta}}}{\theta^n(n-1)!}}$$ $$=$$ $$\frac{\theta^{2n} (n-1)!e^{\frac{1-\theta^2}{\theta}\sum^n_{i=1}x_i}}{{\sum^n_{i=1}}^{n-1}}$$

So by my calculation(probably which are wrong), $$\theta$$ is not cancelling out. Any mistakes that I have made that you could possibly see? Any help would be appreciated!

Here's the theorem in my textbook: • What is your definition of $f_{\theta}^T(T(x))$? Anyway, the natural way to use the Factorisation Theorem is to show that the joint probability density of the sample is the product of a function $g$ of the statistic and the parameter $\theta$, and a function $h$ that does not depend on $\theta$, as is done without difficulty at the bottom of this page: online.stat.psu.edu/stat414/node/283. Jan 20 '20 at 11:43
• No, but I would not like to prove it using Fisher-Neymann factorisation theorem, but this way which is in the question
– user255658
Jan 20 '20 at 12:02
• Ok. How do you define $f_{\theta}^{T}(T(x))$? I am puzzled by the superscript. Jan 20 '20 at 12:04
• @MickyboYakari it is the pdf of $T(X)=\sum^n_{i=1}X_i \sim \Gamma(n,\theta)$
– user255658
Jan 20 '20 at 12:08
• All right. I'll write the answer then. Jan 20 '20 at 12:18

Because the individuals are sampled independently from an exponential distribution with parameter $$\theta$$, we have got $$T(\boldsymbol{X})=\sum_{i=1}^{n} X_i \sim \Gamma(\alpha=n,\beta=\theta).$$ So, $$f_\theta^{T}(T(\boldsymbol{X}))=\theta^n \frac{1}{(n-1)!} (\sum_{i=1}^{n} X_i)^{n-1} e^{-\theta \sum_{i=1}^{n} X_i }$$ and $$\frac{f_{\theta}(x)}{f_\theta^{T}(T(\boldsymbol{X}))} =\frac{ (n-1)!}{\sum_{i=1}^{n} X_i},$$ which is a constant function with respect to $$\theta$$.
• oh, so my only mistake was taking $\beta$ for $1\over \theta$ .
• One should always be wary of distributions which have more than one classical parametrisation. This is the reason I like the $\Gamma(\alpha=\alpha_0,\beta=\beta_0)$ notation. Jan 20 '20 at 13:22