# Let $X_1, X_2, X_3 \sim \textrm{Expo}(1)$ distribution. How to compute $\mathrm{P}(X_1 + X_2 \leq rX_3)$?

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.

• I don't immediately see how the CLT would come into this. Commented Dec 3, 2016 at 7:06
• @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. Commented Dec 4, 2016 at 0:25
• 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 Commented Dec 4, 2016 at 3:14
• @gung What hints can you suggest to provide here? Commented Dec 6, 2016 at 16:46

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 $Z \sim \textrm{Beta}(1,2)$

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

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 $$x = \sqrt{r/(1+r)}$$.