Question

Let \(X_{1}, X_{2}, \ldots, X_{5}\) be iid with pdf \(f(x)=e^{-x}, 0

Short answer

By proving that the joint pdf of \( (X1 + X2) / (X1 + X2 + \ldots + X5) \) and \( (X1 + X2 + \ldots + X5) \) equals the product of their respective pdfs, we have demonstrated that these two variables are indeed statistically independent.

Step-by-step solution

Compute the pdfs of the Variables

We compute the pdf of \(X_1 + X_2\) by adding two random variables and then the pdf of \(X_1 + X_2 + \ldots + X_5\) by adding five random variables. The pdf of the sum of two iid random variables with pdf \(f(x) = e^{-x}\) is given by the convolution of \(f(x)\) with itself, and the pdf of the sum of five iid random variables is given by the convolution of \(f(x)\) with itself four times. In the former case, the result is \(f two_{sum}(x) = \int_0^{x} e^{-u} e^{-(x-u)} du\) and in the latter case, the result is \(f five_{sum}(x) = \int_0^{x} f two_{sum}(u) e^{-(x-u)} du\).

Compute Joint pdf

To compute the joint pdf of the two random variables, we use the definition of the joint pdf of two random variables U and V which are functions of a set of iid random variables X1, X2, ..., Xn, given that the Jacobian of the transformation is nonzero: \(f U,V(u,v) = f X1,X2,...,Xn(x1,x2,...,xn) |J|\), where |J| is the absolute value of the Jacobian of the inverse transformation from U and V to X1, X2, ..., Xn. Here, we take \(U = (X_1 + X_2) / (X_1 + X_2 + \ldots + X_5)\) and \(V = (X_1 + X_2 + \ldots + X_5)\).

Show that variables are Independent

Next, we show that the above pdf can be written as the product of the individual pdfs of U and V. By showing that the joint pdf equals the product of the individual pdfs, we demonstrate that U and V are independent. The mathematical details can be tedious and involve a fair bit of algebra and integration techniques.