Problem Statement: Suppose $Y_1, Y_2,\dots, Y_n$ are a random sample from an exponential distribution with mean $\theta.$ Let $\displaystyle U=\sum_{i=1}^n Y_i.$ Find the conditional density function for $Y_1$ given $U:\; f_{Y_1|U}(y_1|u).$

Note: This is essentially Exercise 9.86b in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Scheaffer. The problem is posed in the context of $U$ as a sufficient statistic, and of using the Rao-Blackwell Theorem to find the MVUE of $e^{-t/\theta}.$

My Work So Far: The conditional density function is given by \begin{align*} f_{Y_1|U}(y_1|u) &=\frac{f(y_1,u)}{f_U(u)}. \end{align*} Now the density function for $Y_1$ is given by $$f_1(y_1)=\frac{1}{\theta}\,e^{-y_1/\theta},$$ and the sum function $U=\sum Y_i$ would then have, by virtue of the mgf, \begin{align*} m_U(t) &=\prod_{i=1}^n(1-\theta t)^{-1}\\ &=(1-\theta t)^{-n}, \end{align*} which is the mgf for a Gamma distribution with parameters $\beta=\theta$ and $\alpha=n.$ It follows that $$f_U(u)=\frac{u^{n-1}\,e^{-u/\theta}}{\Gamma(n)\,\theta^n}.$$ Now $u=y_1+y_2+\cdots+y_n,$ so that $y_1=u-y_2-y_3-\cdots-y_n.$ By the logic we have already used, if we let $S=\sum_{i=2}^nY_i,$ then the distribution function for $S$ is a Gamma with parameters $\beta=\theta$ and $\alpha=n-1.$ Also, $u=y_1+s,$ or $y_1=u-s.$ Note that $y_1$ and $s$ are independent. It follows that \begin{align*} f_{(Y_1,S)}(y_1,s) &=f_{Y_1}(y_1)\,f_S(s)\\ &=\frac1\theta\,e^{-y_1/\theta}\, \frac{s^{n-2}\,e^{-s/\theta}}{\Gamma(n-1)\,\theta^{n-1}}\\ &=\frac{s^{n-2}\,e^{-(y_1+s)/\theta}}{\Gamma(n-1)\,\theta^n}. \end{align*}

My Question: I think I'm on the right track, but I also think I'm missing something fairly straight-forward. How do I push through to get the desired conditional probability density function?

I have examined a few questions, especially this one. I do not understand the method of solution there, and the notation they are using is confusing: are they working with $U$ or $Y_n?$ This question is taking a different approach (doing the expectation without knowing the distribution function), and so doesn't help me.

  • 1
    $\begingroup$ stats.stackexchange.com/questions/252692, which poses the same exercise, presents a different kind of solution. I'm not voting to close, though, because you are really asking about your particular method with an aim to addressing a related problem. You still might find some of the ideas used there--especially concerning the exchangeability of the $Y_i$--will be helpful in general. $\endgroup$
    – whuber
    Commented Jul 14, 2021 at 21:45


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