Finding the mgf of $X_1 - X_{(1)}$ from Basu's Theorem

Question

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.

Answer 1

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

$$E\left[e^{tU}\right]=\frac{e^{t\theta}}{1-t}\cdot\frac{n-t}{ne^{t\theta}}=\frac{n-t}{n(1-t)}$$

Answer 2

I think you can compute the cdf of $U$ directly. Fix $u \geq 0$. Note that $$F_{U}(u)=P(U\leq u)=\int_{\theta}^{\infty}P(U\leq u|X_{(1)}=t)f_{X_{(1)}}(t)dt$$ Let's take a closer look at $P(U \leq u|X_{(1)}=t)$. We can say $$P(U \leq u|X_{(1)}=t)=P(U \leq u|X_{(1)}=t,X_{(1)}=X_1)P(X_{(1)}=X_1|X_{(1)}=t)+P(U \leq u|X_{(1)}=t,X_{(1)}\neq X_1)P(X_{(1)} \neq X_1|X_{(1)}=t)$$ This simplies to $$P(U \leq u|X_{(1)}=t)=1\cdot \frac{1}{n}+ \frac{\int_t^{t+u}f_{X}(x)dx}{\int_t^{\infty}f_{X}(x)dx}\cdot \frac{n-1}{n}=\frac{1+(n-1)(1-e^{-u})}{n}$$ Plugging this into our formula for $F_{U}$ and integrating yields $$F_{U}(u)=\frac{1+(n-1)(1-e^{-u})}{n}\cdot I(u \geq 0)$$ I'm reluctant to utilize the pdf of $U$ in in our calculation of the moment generating function since $U$ isn't technically a continuous random variable; in fact, the event $\{U=0\}$ occurs with probability $1/n$. To get around this, we can compute $M_{U}(t)=\mathbb{E}(e^{tU})$ via the integral $\int_{0}^{\infty}P(e^{tU}>x)dx$. If $t>0$ is fixed, then $e^{tU}$ is supported on $[1,\infty)$ and so $$M_{U}(t)=1+\int_1^{\infty}\bigg[1-F_{U}\big(\ln(x)/t\big)\bigg]dx=1+\frac{n-1}{n}\int_1^{\infty}\frac{dx}{x^{1/t}}$$ The improper integral on the RHS converges iff $0<t<1$ and evaluates to $\frac{t(n-1)}{n(1-t)}$. In other words, $$M_{U}(t)=1+\frac{t(n-1)}{n(1-t)}$$ for $t\in(0,1)$. Proceeding in this manner for $t<0$ yields the same formula $M_{U}(t)$, and clearly $M_{U}(0)=1$. We conclude $M_{U}(t)=1+\frac{t(n-1)}{n(1-t)}$ is our mgf and has domain $(-\infty,1)$.