I'm stuck on an exercise (it's not homework, but preparation for finals). It goes like this: $X_1, \dots, X_n$ are iid Exponential($\lambda$) (with parametrization $f(x)=\lambda e^{-\lambda x}$). What is the pdf $f(x_n|Y)$, where $Y=\sum_{i=1}^n X_i$? I know that $Y\sim Gamma(n, \lambda^{-1})$ with pdf $$ f(y)=\frac{\lambda^n}{\Gamma(n)}y^{n-1}e^{-y\lambda}. $$

So I have the marginals, but I'm not sure how to proceed. Maybe it's easier to get $f(y|X_n)$ and then multiply by the ratio of the marginals? Any help that puts me in the right direction is appreciated!

Edit: I think I solved it. Since all are iid, the joint is simply $f(x_n, t_{n-1})$ where $T_{k}=\sum_{i=1}^kX_i$. This comes from a product of convolutions. The joint of $(x_2, x_1)$ is just the product of the densities. The joint of $$ f(x_3, t_2)=f_{X_3}(x_3)f_{T_2}(t_3-x_3) $$ and so on. So finally $$ f(x_n, t_{n-1})=f_{X_n}(x_n)f_{T_{n-1}}(t_n-x_n). $$ The former is exponential, the latter is gamma, so the conditional is $$ f(x_n|T_n)=\frac{f_{X_n}(x_n)f_{T_{n-1}}(t_n-x_n)}{f_{T_n}(t_n)}. $$

  • $\begingroup$ A subtle point: it appears that the $X$ rv is also included in $Y$. So the joint is $f(x_n,y) = f(x_n, x_n + \sum^{n-1}x_i)$. If not, you should clarify that. $\endgroup$ Dec 21, 2016 at 15:12
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    $\begingroup$ To which of these variables does "$x$" refer to in the expression "$f(x|Y)$"? $\endgroup$
    – whuber
    Dec 21, 2016 at 15:28
  • $\begingroup$ @AlecosPapadopoulos Yes, precisely. But $f(x_n, \sum_{i=1}^nx_i)$ is just the same as $f(x_n, \sum_{i=1}^{n-1}x_i)$; loosely speaking, having the sum up to $n$ shouldn't make a difference as $x_n$ then appears twice. This shouldn't be controversial, but maybe I was vague here. I essentially just used the approach in Sasha's answer here: math.stackexchange.com/questions/335894/…. $\endgroup$
    – jacknick
    Dec 21, 2016 at 19:47
  • $\begingroup$ @whuber: It refers to any of $n$ $x$-variables in the sum. They are all iid, so which specific $i$ we consider is of no importance (rather than perhaps notational convenience). $\endgroup$
    – jacknick
    Dec 21, 2016 at 19:50
  • $\begingroup$ @jacknick $x_n$ is independent from $\sum^{n-1}$, but it is not independent of $\sum^{n}$. Also the marginal distribution of $\sum^{n-1}$ is not the same as that of $\sum^{n}$ (they may be of the same family, but they will have different parameter values). In what sense then the two joint densities are "just the same"? I am missing something here. $\endgroup$ Dec 21, 2016 at 19:53

2 Answers 2


It can be instructional and satisfying to work this out using basic statistical knowledge, rather than just doing the integrals. It turns out that no calculations are needed!

Here's the circle of ideas:

  • The $X_i$ can be thought of as waiting times between random events.

  • When the waiting times have independent identical exponential distributions, the random events are a Poisson process.

  • When normalized by the last time (given by $Y=X_1+X_2+\cdots + X_n$), these events therefore look like $n-1$ independent uniform values in $[0,1]$.

  • The values $0 \le X_1/Y \le X_1/Y+X_2/Y \le \cdots \le (X_1/Y+\cdots+X_{n-1}/Y) \le 1$ therefore are the order statistics for $n-1$ iid uniform variables.

  • The $k^\text{th}$ order statistic has a Beta$(k, n-k)$ distribution.

  • The PDF of a Beta$(k,n-k)$ distribution is proportional to $x^{k-1}(1-x)^{n-k-1}$ for $0\le x \le 1$, with constant of proportionality equal to (of course!) the reciprocal of the Beta function value $B(k,n-k)$.

  • Since $Y$ is invariant under any permutation of the $X_i$ and the $X_i$ are exchangeable, all the conditional distributions $f(X_i|Y)$ are the same.

Thus, the distribution of any of the $X_i$ conditional on $Y$ must be $Y$ times a Beta$(1,n-1)$ distribution. Scaling the Beta PDF by $y$ gives the conditional probability element

$$f_{X_1|Y=y}(x)\, \mathrm{d}x = \frac{1}{B(1,n-1)}\left(1-\frac{x}{y}\right)^{n-2}\frac{\mathrm{d}x}{y}$$

for $0 \le X_i \le y$.

This reasoning further implies the $n$-variate distribution of the $X_i$ conditional on $Y$ is $Y$ times a symmetric Dirichlet distribution.


Balakrishnan, N. and A. Clifford Cohen, Order Statistics and Inference. Academic Press, 1991.

  • $\begingroup$ I am not sure how the answer concludes that the Beta marginals imply a joint distribution which is a Dirichlet (my main question is: will the joint distribution be unique?), and I am not sure how to relate the Dirichlet distribution to order statistics as mentioned in the answer. In this answer (stats.stackexchange.com/questions/36093/…), you talk about a similar question, but I can't wrap my head around how marginals are enough to specify the joint distribution here. $\endgroup$
    – BlackHat18
    Dec 22, 2020 at 14:03
  • $\begingroup$ @Black You may be overthinking this: set $Y=1$ and immediately draw your conclusion. $\endgroup$
    – whuber
    Dec 22, 2020 at 14:06
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    $\begingroup$ @Black Consider editing the question you posted to focus on this issue (which I think may be the heart of the matter). $\endgroup$
    – whuber
    Dec 22, 2020 at 14:12
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    $\begingroup$ @Black Please take a few minutes to review our help center. It will explain how your question goes into a review queue for the community to vote on. $\endgroup$
    – whuber
    Dec 22, 2020 at 14:35
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    $\begingroup$ @Adrian In this case, to emphasize that the variable of integration is $x$ (and not $y$ or $(x,y),$ both of which would otherwise be reasonable interpretations of the notation). In other cases, such as when transforming densities, it ensures the transformation is correctly computed. See our posts referencing probability elements. $\endgroup$
    – whuber
    Jul 15, 2021 at 14:54

This might be the most "text book" answer on $f_{X_1|Y}(x_1|y)$.

Let $Z = X_2 + ... + X_n$. Then $Y = Z + X_1$.

First the joint distribution of $(Z, X_1)$ which is $f_{Z,X_1}(z,x_1) = \dfrac{\lambda^{n-1}}{\Gamma(n-1)}\,z^{n-2}\,e^{-z\lambda}\,\lambda \times e^{-x_1 \lambda}$ for $z \ge 0, x_1 \ge 0$.

Next we get the joint distribution of $(Y, X_1)$, which is $f_{Y,X_1}(y,x_1) = f_{Z,X_1}(z,x_1)$ since the absolute determinant of the Jacobian matrix is 1 (details below).

Therefore, $f_{Y,X_1}(y,x_1) = \dfrac{\lambda^n}{\Gamma(n-1)}\,(y-x_1)^{n-2}\,e^{-y\lambda}$ for $x_1 \ge 0, y \ge x_1$, and $0$ otherwise.

Finally, $f_{X_1|Y}(x_1|y) = \dfrac{f_{Y,X_1}(y,x_1)}{f_Y(y)}$. The rest follows easily from here.

Edit: I didn't see OP's "edit" part at first. So OP has solved it in the same "text book" way. I will leave my post here, in case it should be useful to others.

Edit 2: Details on the Jacobian.

We are doing variable transformations, from $Z, X_1$ to $Y, X_1$, defined as $Y=Z+X_1, X_1 = X_1$. So Jacobian matrix is 2-by-2, with the elements being (going across rows) $\dfrac{\partial Y}{\partial Z}, \dfrac{\partial Y}{\partial X_1}, \dfrac{\partial X_1}{\partial Z}, \dfrac{\partial X_1}{\partial X_1}$, i.e. $1,1,0,1.$ The absolute determinant of this matrix is $1.$

  • $\begingroup$ It would be very illustrative to me if you could please flesh out the Jacobian part of your argument a bit more. Which are the functions you're differentiating? And with respect to which variables? Thanks! $\endgroup$ Jul 15, 2021 at 17:50
  • $\begingroup$ Sure. No problem. See Edit above. $\endgroup$ Jul 16, 2021 at 17:03
  • $\begingroup$ Great, thanks! That's very helpful! $\endgroup$ Jul 16, 2021 at 18:40

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