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So let $X_1, X_2, X_3$ be random samples from the $\textrm{Expo}(1)$ distribution.

How do I set up the computation for

$\mathrm{P}(X_1 + X_2 \leq rX_3 | \sum_{i=1}^n X_i = t)$ where $r, t > 0$?

Edit: I do not know how to use the notion that $X_1 + X_2$ is independent from $\frac{X_1}{X_1 + X_2}$ to proceed with the problem.

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    $\begingroup$ I don't immediately see how the CLT would come into this. $\endgroup$
    – Glen_b
    Commented Dec 3, 2016 at 7:06
  • $\begingroup$ @Glen_b, I thought of CLT, because I saw $X_1, X_2, ... X_n$ , but yes, it might not come into this. Editing to reflect that. $\endgroup$
    – potpie
    Commented Dec 4, 2016 at 0:25
  • $\begingroup$ While an algebraic answer may well be doable here, in a practical situation I'd be inclined to use simulation, especially if it's a one-off computation $\endgroup$
    – Glen_b
    Commented Dec 4, 2016 at 3:14
  • $\begingroup$ @gung What hints can you suggest to provide here? $\endgroup$
    – wolfies
    Commented Dec 6, 2016 at 16:46

2 Answers 2

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So I think here is a way of solving the question analytically:

$\mathrm{P}(X_1 + X_2 \leq rX_3 | \sum_{i=1}^n X_i = t)\\ \rightarrow \mathrm{P}\left(X_1 + X_2 + X_3 \leq (1 + r)X_3 | \sum_{i=1}^n X_i = t\right)\\ \rightarrow \mathrm{P}\left(\frac{X_1 + X_2 + X_3}{X_3} \leq (1 + r) | \sum_{i=1}^n X_i = t\right)\\ \rightarrow \mathrm{P}\left(\frac{X_3}{X_1 + X_2 + X_3} \geq \frac{1}{1 + r} | \sum_{i=1}^n X_i = t\right)\\$

The sum of exponentially distributed r.v.'s follows a Gamma distribution. Hence, we have that

$X_1 + X_2 \sim \textrm{Gamma}(2,1)\\ X_3 \sim \textrm{Gamma}(1,1)$

And so,

$\mathrm{P}\left(\frac{X_3}{X_1 + X_2 + X_3} \geq \frac{1}{1 + r} | \sum_{i=1}^n X_i = t\right)\\ \rightarrow \mathrm{P}\left(\frac{\textrm{Gamma}(1,1)}{\textrm{Gamma}(2,1) + \textrm{Gamma}(1,1)} \geq \frac{1}{1 + r} | \sum_{i=1}^n X_i = t\right)\\ \rightarrow \mathrm{P}\left(\textrm{Beta}(1,2) \geq \frac{1}{1 + r} | \sum_{i=1}^n X_i = t\right)$

We can drop the conditional by noting that

$\frac{X_3}{X_1 + X_2 + X_3} \perp \sum_{i=1}^n X_i$.

$U + V \perp \frac{U}{U+V}$ if $U$ and $V$ are Gamma distributed, and we let $U = X_1 + X_2$ and $V = X_3$.

To compute

$\mathrm{P}\left(\textrm{Beta}(1,2) \geq \frac{1}{1 + r}\right)$

we let a random variable $Z \sim \textrm{Beta}(1,2)$

$\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 \frac{1}{\textrm{Beta}(1,2)}z^{1-1}(1-z)^{2-1}\textrm{d}z\\ \rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 \frac{\Gamma(3)}{\Gamma(1)\Gamma(2)}z^{1-1}(1-z)^{2-1}\textrm{d}z\\ \rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 \frac{2!}{0!1!}z^{1-1}(1-z)^{2-1}\textrm{d}z\\ \rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 2z^{0}(1-z)^{1}\textrm{d}z\\ \rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \int_{\frac{1}{1+r}}^1 2-2z\textrm{d}z\\ \rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = (2z - z^2)|_{\frac{1}{1+r}}^1\\ \rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = 2 - 1 - \frac{2}{1+r} + \frac{1}{(1+r)^2}\\ \rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \frac{-1+r}{1+r} + \frac{1}{(1+r)^2}\\ \rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \frac{r^2 - 1}{(1+r)^2} + \frac{1}{(1+r)^2}\\ \rightarrow\mathrm{P}\left(Z \geq \frac{1}{1 + r}\right) = \frac{r^2}{(1+r)^2}$

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There's a simpler way to solve this. First, note that the joint distribution of the $X_i / \sum X_i$ is Dirichlet$(1,1,1)$. This immediately implies that the distribution of $(X_1 + X_2)/(X_1+X_2+X_3)$ is Beta$(2,1)$. This holds for whatever value $t$ we constrain $\sum X_i$ to equal, so we may as well work with the normalized $X_i$ and their Dirichlet distribution, ignoring $t$ altogether.

We now need to find the probability that a Beta$(2,1)$ variate, label it $z$, is $\leq r*(1-z)$. A small amount of rearranging of the inequality leads to the condition $z \leq r/(1+r)$.

Thus, our calculation involves finding the value of $z$ for which the cumulative density function of $z$ equals $r/(1+r)$. Since the Beta$(2,1)$ distribution has the simple functional form $f_{\beta}(x) = 2x$, we can find an analytic solution quite easily. $F_{\beta}(x) = x^2$, and setting $x^2 = r/(1+r)$ leads to the solution $x = \sqrt{r/(1+r)}$.

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