Solution Found!
Let X1 and X2 be a random sample of size n = 2 from a distribution with pdf f(x) = 6x(1
Chapter 5, Problem 5.3-6(choose chapter or problem)
Let \(X_1\) and \(X_2\) be a random sample of size \(n=2\) from a distribution with pdf \(f(x)=6 x(1-x)\), \(0<x<1\). Find the mean and the variance of \(Y=X_{1}+X_{2}\).
Questions & Answers
QUESTION:
Let \(X_1\) and \(X_2\) be a random sample of size \(n=2\) from a distribution with pdf \(f(x)=6 x(1-x)\), \(0<x<1\). Find the mean and the variance of \(Y=X_{1}+X_{2}\).
ANSWER:Step 1 of 2
From the given data we have.
Given that the \(f(x)=6x(1-x),\quad\ \ \ 0<x<1\)
lets assume \(y=x_{1}+x_{2}\)
\(\begin{aligned}mean\ \mu_{y} & =E(y) \\ & =E\left(x_{1}+x_{2}\right) \\ & =E\left(x_{1}\right)+E\left(x_{2}\right) \\ & =\int_{0}^{1} x_{1} f\left(x_{1}\right) d x+\int_{0}^{1} x_{2} f\left(x_{2}\right) d x \\ & =\int_{0}^{1} x_{1} 6 x_{1}\left(1-x_{1}\right) d x+\int_{0}^{1} x_{2} 6 x_{2}\left(1-x_{2}\right) d x \\ & =\left[\frac{-3}{2} x_{1}^{4}+2 x_{1}^{3}\right]_{0}^{1}+\left[-\frac{3}{2} x_{2}^{4}+2 x_{2}^{3}\right]_{0}^{1} \end{aligned}\)
\(\mu_{y}=1\)