Question

For some constant c, the random variable X has the probability density function:

$f\left(x\right)=\left\{\begin{array}{l}c{x}^4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<x<2\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{otherwise}\end{array}\right.$

Find

1. $E\left[X\right]$and
2. $Var\left(X\right)$
   

Short answer

a. The value of $E\left[X\right]=\frac{5}{3}$

b. $\mathrm{Va}r\left(X\right)=\frac{5}{63}$

Step-by-step solution

Step:1 Given Information (Part a)

The random variable X has the probability density function

$f\left(x\right)=\left\{\begin{array}{l}c{x}^4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<x<2\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{otherwise}\end{array}\right.$

We have to find$E\left[X\right]$

Definition (part a)

The Probability Density Function (PDF) is a probability function that represents the density of a continuous random variable that falls inside a given range of values.

Step:3 Explanation (Part a)

Let's start by determining the c constant. There must be one,

$1={\int }_{\mathrm{ℝ}}f\left(x\right)dx={\int }_0^{2}c{x}^{4}dx={\left.c\left(\frac{x^5}{5}\right)\right|}_0^2=c·\frac{32}{5}$

which implies that $c=\frac{5}{32}$

(a) The mean value is

$$\begin{gathered} E\left(X\right)={\int }_{\mathrm{ℝ}}xf\left(x\right)dx={\int }_0^{2}x·\frac{5}{32}{x}^{4}dx=\frac{5}{32}{\int }_0^2{x}^{5}dx \\ ={\left.\frac{5}{32}\left(\frac{x^6}{6}\right)\right|}_0^2=\frac{5}{32}\times \frac{64}{6}=\frac{5}{3} \end{gathered}$$

Explanation (Part b)

According to the information, We observe that:

$E\left(X^2\right)={\int }_0^2 x^2\cdot \frac{5x^4}{32}dx=\frac{5}{32}{\left(\frac{x^7}{7}\right)}_0^2$

$\frac{5}{32}\left(2^7-0\right)=\frac{5}{32\times 7}\times 128=\frac{20}{7}$

Therefore $Var\left(X\right)$:

localid="1650014593098" $Var\left(X\right)=E\left(x^2\right)-\left(E\right(x){)}^2$

localid="1649597273687" $\Rightarrow Var\left(X\right)=\frac{20}{7}-{\left(\frac{5}{3}\right)}^2=\frac{5}{63}$