Probability — Question 6

Question 6

For $f(x)=6x(1-x)$, $0\le x\le1$, find $\operatorname{Var}(X)$. Use symmetry for the mean.

Question 6 – Solution

Step 1: Use symmetry to find the mean. The density satisfies $$f(1-x)=6(1-x)x=f(x),$$ so it is symmetric about $x=1/2$. Therefore, $$\mu=E[X]=\frac12.$$

Step 2: Compute the second moment. $$\begin{align*} E[X^2] &=\int_0^1x^2f(x)\,dx\\ &=\int_0^1x^2(6x-6x^2)\,dx\\ &=\int_0^1(6x^3-6x^4)\,dx\\ &=\left[\frac32x^4-\frac65x^5\right]_0^1 =\frac32-\frac65=\frac3{10}. \end{align*}$$

Step 3: Apply the variance formula. $$\begin{align*} \operatorname{Var}(X) &=E[X^2]-(E[X])^2\\ &=\frac3{10}-\left(\frac12\right)^2\\ &=\frac3{10}-\frac14=\frac1{20}. \end{align*}$$ $$\boxed{\operatorname{Var}(X)=\frac1{20}}$$