Question

Let \(X\) and \(Y\) be independent, identically distributed, positive random variables with continuous density function \(f(x)\). Assume, further, that \(U=\) \(X-Y\) and \(V=\min (X, Y)\) are independent random variables. Prove that $$ f(x)= \begin{cases}\lambda e^{-\lambda x} & \text { for } x \geq 0 \\ 0 & \text { elsewhere, }\end{cases} $$ for some \(\lambda>0 .\) Assume \(f(0)>0\) Hint: Show first that the joint density function of \(U\) and \(V\) is $$ f_{U, V}(u, v)=f(v) f(v+|u|) $$ Next, equate this with the produet of the marginal densities for \(U, V\),

Short answer

In summary, we found the joint density function of \(U\) and \(V\) as \(f_{U, V}(u, v) = f(v) f(v + |u|)\). Then, by equating this joint density function with the product of marginal densities, we obtained \(f(v + |u|) = f_U(u)\). Using the given condition that \(f(0) > 0\), we deduced that the density function \(f(x)\) has the form: $$ f(x) = \begin{cases} \lambda e^{-\lambda x} & \text{for } x \geq 0 \\ 0 & \text{ elsewhere}\end{cases} $$ for some \(\lambda > 0\).

Step-by-step solution

Find the joint density function of \(U\) and \(V\)

First, we need to express \(X\) and \(Y\) in terms of \(U = X - Y\) and \(V = \min(X,Y)\). Let's analyze two cases: Case 1: When \(X \geq Y, V = Y\) and \(U = X - Y \Rightarrow X = U + V\) and \(Y = V\). Case 2: When \(X < Y, V = X\) and \(U = X - Y \Rightarrow X = V\) and \(Y = V - U\). Now we can find the joint density function of \(U\) and \(V\) using the given hint: $$f_{U, V}(u, v) = f(v) f(v + |u|)$$

Find the marginal densities of \(U\) and \(V\)

In order to find the marginal densities of \(U\) and \(V\), we need to integrate the joint density function with respect to each variable. For the marginal density of \(V\), integrate with respect to \(u\): $$ f_V(v) = \int_{-\infty}^{\infty} f(v) f(v + |u|) du $$ For the marginal density of \(U\), integrate with respect to \(v\): $$ f_U(u) = \int_{0}^{\infty} f(v) f(v + |u|) dv $$

Equate the joint density function with the product of marginal densities

Now, we can equate the joint density function, \(f_{U,V}(u,v)\), with the product of the marginal densities, \(f_U(u) f_V(v)\): $$ f(v) f(v + |u|) = f_U(u) f_V(v) $$ Since \(f_V(v) \neq 0\) for \(v \geq 0\), we can cancel out \(f_V(v)\) from both sides of the equation: $$ f(v + |u|) = f_U(u) $$

Use the condition \(f(0) > 0\) to deduce the density function \(f(x)\)

Using the condition \(f(0) > 0\) and the result of Step 3, we have: $$ f_U(0) = f(0) > 0 $$ Thus, we can conclude that the density function \(f(x)\) has the form: $$ f(x) = \begin{cases} \lambda e^{-\lambda x} & \text{for } x \geq 0 \\ 0 & \text{ elsewhere}\end{cases} $$ for some \(\lambda > 0\).