# Sufficiency in the exponential distribution

I am trying to show that given a random sample $$\{X_i\}_{i=1}^n$$ where $$X_i\sim exp(\lambda^{-1})$$, the statistic $$T(\mathbf{X})=\sum_{i=1}^n X_i$$ is sufficient by using only the definition.

I have the following:

$$f_{\mathbf{X}}(\mathbf x) = f_{\mathbf{X},T(\mathbf{X})}(\mathbf x,t) = \begin{cases} \lambda^{-n}e^{-\lambda^{-1}t}, & \text{when } t=\sum_{i=1}^nX_i\\ 0, & \text{otherwise} \end{cases}$$ By using the fact that $$T(\mathbf X)\sim \Gamma(n,\lambda^{-1}),$$ $$f_{T(\mathbf{X})}(t)=\frac{\lambda^{-n}t^{n-1}e^{-\lambda^{-1}t}}{\Gamma(n)}$$

Now when I calculate the conditional distribution, I get

$$f_{\mathbf X| T(\mathbf X)=t}(\mathbf x)=\frac{\Gamma(n)}{t^n-1}$$

However, I am not sure about the support of the last expression (so that it integrates to 1). I believe it has to be something like this:

$$0, $$s_1, $$s_2, .... , $$s_{n-2} where we define $$s_j=\sum_{i=1}^j x_i$$ and the variable $$x_n$$ is omitted because it is linearly dependent of the other ones.

Can anyone comment on whether there is a mistake in my line of thought?

• Lazy notation, just fixed it. – Daniel Ordoñez Apr 17 at 15:12