Question

For 0 ≤ x ≤ 1 let f(x) = kx(1 − x), where k is a constant. find the value of k such that f is a density function. 2. find the mean and variance of the distribution in the preceding exercise.gle.com/search?q=.+find+the+mean+and+variance+of+the+distribution+in+the+preceding+exercise.&oq=.+find+the+mean+and+variance+of+the+distribution+in+the+preceding+exercise.&aqs=chrome..69i57.485j0j7&sourceid=chrome&ie=utf-8

Answer 1

For $f_X(x)$ to be a proper density function, we need

$\displaystyle\int_{-\infty}^\infty f_X(x)\,\mathrm dx=1$

We have

$\displaystyle\int_{-\infty}^\infty f_X(x)\,\mathrm dx=k\int_0^1x(1-x)\,\mathrm dx=\frac k6=1\implies k=6$

The mean is given by

$\mathbb E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=6\int_0^1x^2(1-x)\,\mathrm dx=\frac12$

The variance is

$\mathbb V[X]=\mathbb E[X^2]-\mathbb E[X]^2$

where

$\mathbb E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=6\int_0^1x^3(1-x)\,\mathrm dx=\frac3{10}$

so that

$\mathbb V[X]=\dfrac3{10}-\dfrac1{2^2}=\dfrac1{20}$