Suppose that $X_1, \ldots, X_n$ where $n \geq 2$ are iid random variables with pdf
$$f_X(x) = e^{-(x-\theta)}I(x \geq \theta).$$
I need to find the moment-generating function of the statistic $U = X_1 – X_{(1)}$ where $X_{(1)}$ is the sample minimum. Prior to this, I have found that the distribution function of $T = X_{(1)}$ is $$
f_T(t) = ne^{-n(t-\theta)}I(t\geq\theta).$$
I have also shown that $X_{(1)}$ is complete sufficient and that $U$ is ancillary, and as such, by Basu's theorem, I know that these two statistics are independent.
Now I am supposed to use this information to find the mgf and cdf of $U$. Since $U$ is bounded I know that I can determine the moment-generating function by taking
$$E[U^k] = E[(X_1 – X_{(1)})^k]$$
rather than directly computing the moment generating function (since I do not know the distribution of $U$, just that $U$ is ancillary w.r.t. $\theta$). However, I do not know how to compute this expectation.
I was hoping there would be an algebraic trick that would allow me to use the independence of $T$ and $U$ in computing this expectation, but I cannot find one (I believe that this is how the exercise is intended to be solved, but I am stumped). I did try writing
$$E[U^k] = E\left[\sum_{j=1}^k \binom{k}{j} X_1^k (-X_{(1)})^{k-j}\right] = \sum_{j=1}^k \binom{k}{j} E\left(X_1^j)\right)E\left(-X_{(1)}^{k-j}\right) $$
but this has not helped me compute the expectation, as I do not think that $E(X_1^j)$ has an elementary integral.
If anyone can help me compute this expectation or figure out what trick to use, it would be appreciated.
Best Answer
Due to independence of $U=X_1-X_{(1)}$ and $X_{(1)}$ by Basu's theorem, whenever the expectations exist,
\begin{align} E\left[e^{tX_1}\right]&=E\left[e^{t\left(U+X_{(1)}\right)}\right] \\&=E\left[e^{tU}\right]E\left[e^{tX_{(1)}}\right] \end{align}
So provided MGF of both $X_1$ and $X_{(1)}$ exist and $E\left[e^{tX_{(1)}}\right]\ne 0$, we must have
$$E\left[e^{tU}\right]=\frac{E\left[e^{tX_1}\right]}{E\left[e^{tX_{(1)}}\right]}$$
As $(X_i-\theta)$'s are i.i.d $\text{Exp}(1)$, distribution of $X_{(1)}-\theta$ is exponential with mean $\frac1n$.
This means $$E\left[e^{t\left(X_1-\theta\right)}\right]=\frac1{1-t}\quad, \, t<1$$
and $$E\left[e^{t\left(X_{(1)}-\theta\right)}\right]=\frac{n}{n-t}\quad, \, t<n$$
Hence MGF of $U$ for $t<1$ is
$$E\left[e^{tU}\right]=\frac{e^{t\theta}}{1-t}\cdot\frac{n-t}{ne^{t\theta}}=\frac{n-t}{n(1-t)}$$