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??
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*}
