Probability Theory – Problem 14: Find the distributions of and

Let $X_1 ,X_2 \overset{\text{i.i.d.}}{\sim} \mathcal{E}(1)$, and let $\displaystyle U = \frac{X_1}{X_1 + X_2},V = X_1+X_2$. Find the distributions of $U$ and $V$. Furthermore, determine whether $U$ and $X_1$ are independent.

Given that $X_1, X_2$ are independent and identically distributed as the exponential distribution $\mathcal{E}(1)$, their probability density functions are $$f_{X_1}(x) = e^{-x}, \quad x > 0$$ $$f_{X_2}(y) = e^{-y}, \quad y > 0$$ Since $X_1, X_2$ are mutually independent, their joint probability density function is $$f_{X_1, X_2}(x_1, x_2) = f_{X_1}(x_1) f_{X_2}(x_2) = e^{-x_1} e^{-x_2} = e^{-(x_1+x_2)}$$ where $x_1 > 0, x_2 > 0$. The transformation is given by: $$u = \frac{x_1}{x_1 + x_2}$$ $$v = x_1 + x_2$$ We need to find the inverse transformation, i.e., express $x_1, x_2$ in terms of $u, v$: From $v = x_1 + x_2$ and $u = \frac{x_1}{v}$, we obtain $x_1 = uv$. Solving for $x_2$, we get $x_2 = v - x_1 = v - uv = v(1-u)$. The inverse transformation is $x_1 = uv, x_2 = v(1-u)$.

Since $x_1 > 0$ and $x_2 > 0$, we have $v = x_1 + x_2 > 0$. Moreover, $u = \frac{x_1}{x_1 + x_2}$; since $x_1 > 0$ and $x_1+x_2 > x_1$, it follows that $0 < u < 1$. Therefore, the support of the new variables $(U, V)$ is $\{(u,v) \mid 0 < u < 1, v > 0\}$. We compute the Jacobian determinant $J$: $$J = \det \begin{pmatrix} \frac{\partial x_1}{\partial u} & \frac{\partial x_1}{\partial v} \\ \frac{\partial x_2}{\partial u} & \frac{\partial x_2}{\partial v} \end{pmatrix} = \det \begin{pmatrix} v & u \\ -v & 1-u \end{pmatrix} = v(1-u) - u(-v) = v - uv + uv = v$$ Since $v > 0$, the absolute value of the Jacobian is $|J| = v$.

By the change-of-variables formula for multivariate densities, $f_{U,V}(u,v) = f_{X_1,X_2}(x_1(u,v), x_2(u,v)) |J|$, we obtain the joint density of $(U,V)$: $$f_{U,V}(u,v) = e^{-(uv + v(1-u))} \cdot v = e^{-v} \cdot v$$ defined on $0 < u < 1, v > 0$.

For $U$, we integrate the joint density over $v$ on $(0, \infty)$: $$f_U(u) = \int_0^\infty f_{U,V}(u,v) dv = \int_0^\infty v e^{-v} dv$$ This integral equals the Gamma function value $\Gamma(2) = (2-1)! = 1$. Hence $f_U(u) = 1$ for $0 < u < 1$. This shows that $U$ follows the uniform distribution on $(0,1)$.

For $V$, we integrate the joint density over $u$ on $(0, 1)$: $$f_V(v) = \int_0^1 f_{U,V}(u,v) du = \int_0^1 v e^{-v} du = v e^{-v} \int_0^1 1 du = v e^{-v}$$ for $v > 0$. This is the probability density function of the Gamma distribution with shape parameter $\alpha=2$ and scale parameter $\beta=1$.

A necessary and sufficient condition for independence of two random variables is that the conditional expectation of one given the other is constant (or equivalently, that the conditional distribution does not depend on the conditioning variable). We compute the conditional expectation of $U$ given $X_1=x_1$: $$\mathbb{E}[U \mid X_1=x_1] = \mathbb{E}\left[\frac{X_1}{X_1+X_2} \;\middle|\; X_1=x_1\right] = \mathbb{E}\left[\frac{x_1}{x_1+X_2}\right]$$ Since $X_2$ is a random variable, this expectation is computed with respect to the distribution of $X_2$: $$\mathbb{E}\left[\frac{x_1}{x_1+X_2}\right] = \int_0^\infty \frac{x_1}{x_1+x_2} f_{X_2}(x_2) dx_2 = \int_0^\infty \frac{x_1}{x_1+x_2} e^{-x_2} dx_2$$ The result of this integral is a function that depends on $x_1$, rather than a constant. For instance, as $x_1 \to 0^+$, the expectation tends to 0; as $x_1 \to \infty$, the expectation tends to 1. Since the conditional expectation $\mathbb{E}[U \mid X_1=x_1]$ depends on $x_1$, $U$ and $X_1$ are not independent.

The distribution of $U$ is the uniform distribution on $(0,1)$, i.e., $U \sim \mathcal{U}(0,1)$, with probability density function $f_U(u) = 1, \quad 0 < u < 1$. The distribution of $V$ is the Gamma distribution with parameters $\alpha=2, \beta=1$, i.e., $V \sim \Gamma(2,1)$, with probability density function $f_V(v) = v e^{-v}, \quad v > 0$. $U$ and $X_1$ are not independent.