Let $X_1, X_2, \dotsc , X_n$ iid random sample of size $n$ from an exponential distribution with mean $1/λ$. and $S$ is sum of $X_i$. Find the joint conditional pdf of $X_1, X_2, \ldots , X_n$ given $S$

i think $S\sim\operatorname{Gamma}(n, λ)$ and joint conditional pdf $=f(x_1,x_2,x_3,\dotsc, x_n,S)/f_s(S)$

I think $f_s(S)$ is Gamma distribution's pdf but i can't calculate joint distribution pdf $f(x_1,x_2,x_3,\dotsc,x_n,S)$ because $S$ is sum of $X_i$. very confusing

How can solve this??

  • $\begingroup$ You've got $f_s(S)$ where you should have $f_S(s).$ $\endgroup$ – Michael Hardy Oct 10 '17 at 17:39
  • $\begingroup$ You are right about the distribution of the sum. $\endgroup$ – Michael Hardy Oct 10 '17 at 17:40
  • $\begingroup$ Since $S$ is a function of $(x_1,\ldots,x_n)$, obtain $f(x_1,\ldots,x_n,S) = f(x_1,\ldots,x_n)$. $\endgroup$ – madprob Oct 10 '17 at 17:44
  • $\begingroup$ @MichaelHardy yes, but how can i calculate joint pdf about $x_1,x_2,...x_n,S$ ?? $\endgroup$ – storyable Oct 10 '17 at 17:44
  • $\begingroup$ @madprob ah, really?? why?? $\endgroup$ – storyable Oct 10 '17 at 17:45

This solution is handwavy, since it doesn't explain why $(1) f(x_1,\ldots,x_n,S)=f(x_1,\ldots,x_n)I(S(x_1,\ldots,x_n)=S)$. \begin{align*} f_{X_1,\ldots,X_n|S}(x_1,\ldots,x_n|S) &= \frac{f_{X_1,\ldots,X_n,S}(x_1,\ldots,x_n,S)}{f_S(S)} \\ &= \frac{f_{X_1,\ldots,X_n}(x_1,\ldots,x_n)I(S(x_1,\ldots,x_n)=S)}{f_S(S)} & (1) \\ &= \frac{\prod_{i=1}^{n}{f_{X_i}(x_i)I(n\bar{x}=S)}}{f_{S}(S)} & n\bar{x} = \sum_{i=1}^{n}{x_i} \\ &= \frac{\prod_{i=1}^n \lambda \exp(-\lambda x_i)I(n\bar{x}=S)} {\frac{\lambda^n}{\Gamma(n)}\exp(-\lambda S)} \\ &= \frac{\lambda^n \exp(-\lambda n\bar{x})I(n\bar{x}=S)} {\frac{\lambda^n}{\Gamma(n)}\exp(-\lambda S)} \\ &= \Gamma(n) I(n\bar{x}=S) & \frac{\exp(-\lambda n\bar{x})I(n\bar{x}=S)}{\exp(-\lambda S)}=I(n\bar{x}=S) \end{align*}

  • $\begingroup$ Note that $X_1,\ldots,X_n$ follows a Dirichlet$(1,\ldots,1)$ given that $S=1$. $\endgroup$ – madprob Oct 10 '17 at 19:45
  • $\begingroup$ The original poster uses the capital $S$ as the name of a random variable. Yet you're using it both in that way and as an argument to the density function. $\endgroup$ – Michael Hardy Oct 11 '17 at 0:55
  • $\begingroup$ @MichaelHardy indeed, a conditional probability is a random variable that is defined as $P(X \leq x|S) = E[I(X \leq x)|S]$. Note that a conditional probability is a function of $S$. Similarly, the conditional density is also a random variable such that $f(x|S) = \nabla P(X \leq x|S)$. Unless there is something that I don't see, this part is rigorous. $\endgroup$ – madprob Oct 11 '17 at 1:44
  • $\begingroup$ But you wrote $f(x_1,\ldots,x_n,S). \qquad$ $\endgroup$ – Michael Hardy Oct 11 '17 at 2:44

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