Question

Let \(X_{1}, X_{2}, X_{3}, X_{4}\) be four iid random variables having the same pdf \(f(x)=\) \(2 x, 0

Short answer

The mean and variance of the sum \(Y\) of the four random variables \(X_{1}, X_{2}, X_{3}, X_{4}\) are \(8/3\) and \(2/9\) respectively.

Step-by-step solution

Calculating the Mean (Expectation) of \(X_i\)

The expectation of \(X_i\) is given by the formula \(E[X_i] = \int_{-\infty}^{\infty} x f(x) dx \). By substituting \(f(x) = 2x\) for \(0<x<1\) into the formula, we get: \(E[X_i] = \int_0^1 x * 2x dx = \int_0^1 2x^2 dx = 2/3 .\) Therefore, the expectation of \(X_i\) is \(2/3\).

Calculating the Variance of \(X_i\)

The variance of \(X_i\) is given by the formula \(Var[X_i] = E[X_i^2] - (E[X_i])^2\). To find \(E[X_i^2]\), apply the formula \(E[X_i^2] = \int_{-\infty}^{\infty} x^2 f(x) dx\). By substituting \(f(x) = 2x\) for \(0<x<1\) into the formula, we get: \(E[X_i^2] = \int_0^1 x^2 * 2x dx = \int_0^1 2x^3 dx = 1/2\). Now we can apply these to the variance formula: \(Var[X_i] = E[X_i^2] - (E[X_i])^2 = 1/2 - (2/3)^2 = 1/18 .\) Therefore, the variance of \(X_i\) is \(1/18\).

Calculating the Mean and Variance of \(Y\)

Since \(X_1, X_2, X_3, X_4\) are iid random variables, the mean of their sum \(Y\) is the sum of their means and the variance of \(Y\) is the sum of their variances. \nFor \(Y = X_1 + X_2 + X_3 + X_4\): \nMean of \(Y\), \(E[Y] = E[X_1] + E[X_2] + E[X_3] + E[X_4] = 4*E[X_i] = 4*(2/3) = 8/3\), \nVariance of \(Y\), \(Var[Y] = Var[X_1] + Var[X_2] + Var[X_3] + Var[X_4] = 4*Var[X_i] = 4*(1/18) = 2/9\).