Question

The random variable $X$has the probability density function

$f\left(x\right)=\left\{\begin{array}{ll}ax+bx2& 0<x<1\\ 0& otherwise\end{array}\right.$

If $E\left[X\right]=6$, find

(a) $P\{X<\frac{1}{2}\}$and

(b) $Var\left(X\right)$.

Short answer

(a) The value of $P\{X<\frac{1}{2}\}$is $\frac{7}{20}$

(b) The value of$Var\left(X\right)$is$0.06$

Step-by-step solution

Find the value of P{X<12} (part a)

First and foremost, the density function must satisfy the following conditions:

$$\begin{gathered} 1={\int }_{\mathrm{ℝ}}f\left(x\right)dx \\ ={\int }_0^1\left(ax+b{x}^2\right)dx \\ =a\times \frac{x^2}{2}+{\left.b\times \frac{x^3}{3}\right|}_0^1 \\ =\frac{a}{2}+\frac{b}{3} \end{gathered}$$

We can deduce from the fact that $E\left(X\right)=0.6$,

$$\begin{gathered} 0.6=E\left(X\right) \\ ={\int }_{\mathrm{ℝ}}xf\left(x\right)dx \\ ={\int }_0^1\left(a{x}^2+b{x}^3\right)dx \\ =a\times \frac{x^3}{3}+{\left.b\times \frac{x^4}{4}\right|}_0^1 \\ =\frac{a}{3}+\frac{b}{4} \end{gathered}$$

As a result, we have two unknowns and two linear equations to solve.

$\left\{\begin{array}{ll}3a+2b& =6\\ 20a+15b& =36\end{array}\right.$

$a=\frac{18}{5}$ and $b=-\frac{12}{5}$are the results of this system.

$$\begin{gathered} P\left(X<\frac{1}{2}\right)={\int }_0^{1/2}\frac{18}{5}x-\frac{12}{5}{x}^{2}dx \\ =\frac{9}{5}{x}^2-{\left.\frac{4}{5}{x}^3\right|}_0^{1/2} \\ =\frac{7}{20} \end{gathered}$$

Find Var(X) (part b)

Let's figure out the second moment in order to calculate the variance.

$$\begin{gathered} E\left(X^2\right)={\int }_{\mathrm{ℝ}}{x}^{2}f\left(x\right)dx \\ ={\int }_0^1\frac{18}{5}{x}^3-\frac{12}{5}{x}^{4}dx \\ =\frac{18}{5}\frac{x^4}{4}-{\left.\frac{12}{5}\frac{x^5}{5}\right|}_0^1 \\ =\frac{21}{50} \end{gathered}$$

As a result, we may say that the variance is equal to

$$\begin{gathered} Var\left(X\right)=E\left(X^2\right)-E(X{)}^2 \\ =\frac{21}{50}-0.6^2 \\ =0.06 \end{gathered}$$