# Show that two random variables are independent

I'm learning probability and need help with the following problem :

Let $X_1, X_2$ be independent and identically distributed random variables with probability density function

$$f(x_i) = \begin{cases} \lambda_i e^{\lambda_i x_i}, & x_i > 0 \;\; (i = 1, 2) \\ 0 & \text{elsewhere}.\end{cases}$$

$(1)$ Find the distribution of the random variable $V = \min(X_1, X_2)$.

$(2)$ Show that the random variables $Z = X_1/(X_1+X_2)$ and $X_1+X_2$ are independent.

Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.

$(1)$ If two variables are iid, then they must have the same distribution. Which means they have the same parameters. So I assumed $\lambda_1 = \lambda_2$, i.e.

$$X_1, X_2 \overset{idd}\sim exp(rate = \lambda).$$

To find the distribution of the minimum, we look at cdfs. The cdf for the individual $X_i$’s is

$$F(x) = \int_{-\infty}^{x} f(u) \, du = \int_{0}^{x} f(u) \, du = \int_{0}^{x} \lambda e^{-\lambda u} \, du = 1 - e^{-\lambda x}.$$

The cdf for $V$, the minimum, is :

\begin{align*} F_V(v) = P(V \leq v) &= P(\min(X_1, X_2) \leq v) \\ \\ &= 1 - P(\min(X_1, X_2) > v) \\ \\ &\overset{indep}= 1 - P(X_1 > v) \cdot P(X_2 > v) \\ \\ &\overset{ident}= 1 - [P(X_1 > v)]^n. \end{align*}

From are computed cdf, we have that

$$P(X_1 > v) = 1 - P(X_1 \leq v) = 1 - (1 - e^{-\lambda v}) = e^{-\lambda v}.$$

So, the cdf for the minimum $V$ is

$$F_V(v) = 1 - [e^{-\lambda v}]^2 = 1 - e^{-\lambda 2 v}.$$

The pdf for the minimum is

$$f_V(v) = F'_V(v) = 2\lambda e^{-2\lambda v}.$$

So we proved that $$V \sim exp(rate = 2\lambda).$$

Is my work correct for $(1)$? Unfortunately I have no idea how to solve $(2)$. Any help would be appreciated.

• See an answer here: math.stackexchange.com/a/480323 – NCh Jun 20 '17 at 22:47
• In the question it has explicitly indexed the parameter as $\lambda_1, \lambda_2$ so in general they are not equal. $X_1, X_2$ are independent, both are exponentially distributed, but having different parameter and thus not identically distributed. – BGM Jun 22 '17 at 16:59
• Your work for (1) is correct only for the case $\lambda_1 = \lambda_2 = \lambda$. Can you generalize to the case when $\lambda_1 \neq \lambda_2$? Hint: Look at the complementary CDF $1-F_{\min(X,Y)}(z)=P(X>z, Y>z)$; it is easier to work with. – Dilip Sarwate Sep 13 '17 at 16:00

$(1)$ The variables $X_1$, $X_2\sim \mathcal{E}(\lambda)$ are i.i.d. exponentially distributed. As you showed via the distribution function method, the minimum is exponentially distributed too, and the parameter for the minimum is $2\lambda$, i.e. the sum of the parameters of $X_1$ and $X_2$ (see e.g. [1]).
$(2)$ Since $X_1$, $X_2$ are independent, the joint distribution probability distribution function (PDF) of the vector $X = (X_1, X_2)$ writes: $$f_X(x) = \lambda^2 \exp\left(-\lambda (x_1 + x_2)\right)\,\boldsymbol{1}_{x_1\geq 0,x_2\geq 0} \, .$$ To show that the components $Y_1 = X_1/(X_1 + X_2)$ and $Y_2 = X_1 + X_2$ of the vector $Y = g(X)$ are independent, one can show that the joint PDF $f_Y$ of $Y$ is separable. To do so, let us compute $f_Y$ by the change of variable method. We consider a bounded measurable function $h$, and compute $\mathbb{E}h(Y)$: \begin{aligned} \mathbb{E}(h\circ g)(X) &= \int_{\mathbb{R}^4} h\circ g(x)\, f_{X}(x) \,\mathrm{d}x \, ,\\ &= \int_{\mathbb{R}^4} h(y)\, \underbrace{f_{X}\big({g}^{-1}(y)\big) \big|J_{{g}^{-1}}(y)\big|}_{f_Y(y)} \,\mathrm{d}y \, , \end{aligned} where $J_{g^{-1}}$ is the Jacobian of the reciprocal function of $g$, defined by $$g^{-1}(y) = \big(y_1 y_2 , y_2 (1-y_1)\big) .$$ We get $$f_Y(y) = \lambda^2\, {y_2} \exp\left(-\lambda y_2 \right)\, \boldsymbol{1}_{0\leq y_2}\, \boldsymbol{1}_{0\leq y_1\leq 1} \, ,$$ which is in separated form. Note that this question is similar to [2, 3, 4].