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:

$0<x_1<t$, $s_1<x_2<t$, $s_2<x_3<t$, .... , $s_{n-2}<x_{n-1}<t$ 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?

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

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