# Finding $P(X_{(2)} > 2X_{(1)})$ given $f_{X_{(1)},X_{(2)}}(x_1,x_2)=2e^{-x_1}e^{-x_2}I_{0 \lt x_1 \lt x_2}$

Suppose that $X_1$ and $X_2$ are iid exponential random variables having mean $1$. Give the value of $P(X_{(2)} > 2X_{(1)}$).

Note that $X_{(1)}$ is the sample minimum and $X_{(2)}$ is the second smallest element in the sample, but in this case, the maximum. I'm given that the joint pdf of $X_{(1)}$ and $X_{(2)}$ is

$$f_{X_{(1)},X_{(2)}}(x_1,x_2) = \begin{cases} {2e^{-x_1}e^{-x_2}} & \text{0 \lt x_1 \lt x_2} \\{0} & \text{otherwise}\end{cases}$$

So $x_2$ must be between $2x_1$ and $\infty$ and $x_1\in(0,\infty)$. Thus I have

\begin{align*} P(X_{(2)} > 2X_{(1)}) &= \int_0^\infty\int_{2x_1}^{\infty}2e^{-x_1}e^{-x_2}dx_2 dx_1\\\\ &= \int_0^\infty 2e^{-x_1} \left(-e^{-x_2}|_{2x_1}^{\infty} \right)dx_1 \\\\ &= \int_0^\infty 2e^{-x_1} \left(0-(-e^{-2x_1} \right)dx_1 \\\\ &= \int_0^\infty 2e^{-3x_1}dx_1 \\\\ &=\left(-\frac{2}{3}e^{-3x_1}|_0^{\infty}\right)\\\\ &=0-\left(-\frac{2}{3}\right)\\\\ &= \frac{2}{3}\\\\ \end{align*}

Is this a valid solution? How could one derive the given joint pdf if it were not given?

## 1 Answer

The joint PDF of order statistics has a closed form in terms of CDF and PDF. Here are some links you might look at

To verify your answer, perhaps you know how to calculate the (individual) distributions of $X_{(1)}$ and $X_{(2)}$ using a CDF argument? You should get $$P(X_{(1)} \ge x) = e^{-2x} \text{ i.e. } f_{X_{(2)}}(x) = 2 e^{-2x} \cdot \mathbf{1}[x \ge 0] \text{ i.e. } X_{(1)} \sim \text{Exp}(2)$$ and $$P(X_{(2)} \le x) = (1-e^{-x})^2 \text{ i.e. } f_{X_{(2)}} = (2e^{-x} -2e^{-2x}) \cdot \mathbf{1}[x \ge 0]$$ Therefore \begin{align*} P(X_{(2)} > 2X_{(1)}) &= \int_0^\infty P(X_{(2)} \ge 2X_{(1)} \mid X_{(1)} = x) \cdot f_{X_{(1)}} (x) \; dx \\ &= \int_0^\infty P(X_{(2)} \ge 2x \mid X_{(1)} = x)\cdot (2 e^{-2x}) \; dx \\ &= \int_0^\infty P(X_{(2)} \ge 2x)\cdot (2 e^{-2x}) \; dx \quad \text{(Independent)} \\ &= \int_0^\infty ( 1 - (1-e^{-2x})^2 ) \cdot (2e^{-2x}) \; dx \\ &= \int_0^\infty 4e^{-4x} -2e^{-6x} \; dx \\ &= \left(\int_0^\infty 4e^{-4x} \; dx \right)- \frac{1}{3}\left( \int_0^\infty 6e^{-6x} \; dx \right) \\ &= 1 - \frac{1}{3} = \frac{2}{3} \end{align*}

• Why is $P(X_{(1)}\geq x)=e^{-2x}$? – Remy Nov 26 '17 at 0:29
• yes, $P(X_{(1)} \ge x) = P(X_1 \ge x \text{ and } X_2 \ge x) \stackrel{\text{ind}}{=} P(X_1 \ge x) P(X_2 \ge x) = e^{-x} e^{-x} = e^{-2x}$. Another useful fact in particular for exponential distributions: for independent random variables $X \sim \text{Exp}(a)$ and $Y \sim \text{Exp}(b)$, we have $\min \{ X,Y\} \sim \text{Exp}(a+b)$. (On the other hand, the maximum of independent exponential random variables is not so "nice" and is certainly not exponential) – user365239 Nov 26 '17 at 0:34