Subject: Applied Mathematics 4

Topic: Probability

Difficulty: Medium

$f(x) = k (x−x^2 ) , 0 \leq x \leq 1. k \gt 0$ = 0 ,otherwise

For continuous random variable

$$\int_{\infty}^{\infty} f(x)=1$$

For given problem

$$\begin {aligned}\int_{0}^{1} f(x) d x=1\\ \therefore \int_{0}^{1} k\left(x-x^{2}\right)=1\\ \therefore k\left(\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)_{0}^{1}=1\\ \therefore k\left(\frac{1}{2}- \frac{1}{3}\right)=1\\ \therefore k\left(\frac{3-2}{6}\right)=1\\ \therefore k=6 \end{aligned}$$

Mean

$$\begin{aligned} E(X) &=\int_{\infty}^{-\infty} x f(x) d x\\ &=\int_{0}^{1} x k\left(x-x^{2}\right) d x\\ &=k \int_{0}^{1}\left(x^{2}-x^{3}\right) d x\\ &=k\left(\frac{x^{3}}{3} - \frac{x^{4}}{4}\right)_{0}^{1}\\ &=k\left(\frac{1}{3}-\frac{1}{4}\right)\\ &=6\left(\frac{4-3}{12}\right)\\ &=\frac{1}{2}=0.5 \end{aligned}$$

$$\begin{aligned}E\left(X^{2}\right) &=\int_{\infty}^{-\infty} x^{2} f(x) d x\\ &=\int_{0}^{1} x^{2} k\left(x-x^{2}\right) d x\\ &=\int_{0}^{1}k\left(x^{3}-x^{4}\right) d x\\ &=k\left(\frac{x^{4}}{4}-\frac{x^{5}}{5}\right)_{0}^{1}\\ &=6\left(\frac{1}{4}-\frac{1}{5}\right)\\ &=6\left(\frac{5 - 4}{20}\right)\\ &=\frac{6}{20} \end{aligned}$$

Variance

$$\begin{aligned} Var(X)&=E(X^2)-(E(X))^2 \\ &=\frac{6}{20}-\left(\frac{1}{2}\right)^2\\ &=0.3-0.25\\ &=0.05 \end{aligned}$$

For continuous random variable

$$\int_{\infty}^{\infty} f(x)=1$$

For given problem

$$\begin {aligned}\int_{0}^{1} f(x) d x=1\\ \therefore \int_{0}^{1} k\left(x-x^{2}\right)=1\\ \therefore k\left(\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)_{0}^{1}=1\\ \therefore k\left(\frac{1}{2}- \frac{1}{3}\right)=1\\ \therefore k\left(\frac{3-2}{6}\right)=1\\ \therefore k=6 \end{aligned}$$

Mean

$$\begin{aligned} E(X) &=\int_{\infty}^{-\infty} x f(x) d x\\ &=\int_{0}^{1} x k\left(x-x^{2}\right) d x\\ &=k \int_{0}^{1}\left(x^{2}-x^{3}\right) d x\\ &=k\left(\frac{x^{3}}{3} - \frac{x^{4}}{4}\right)_{0}^{1}\\ &=k\left(\frac{1}{3}-\frac{1}{4}\right)\\ &=6\left(\frac{4-3}{12}\right)\\ &=\frac{1}{2}=0.5 \end{aligned}$$

$$\begin{aligned}E\left(X^{2}\right) &=\int_{\infty}^{-\infty} x^{2} f(x) d x\\ &=\int_{0}^{1} x^{2} k\left(x-x^{2}\right) d x\\ &=\int_{0}^{1}k\left(x^{3}-x^{4}\right) d x\\ &=k\left(\frac{x^{4}}{4}-\frac{x^{5}}{5}\right)_{0}^{1}\\ &=6\left(\frac{1}{4}-\frac{1}{5}\right)\\ &=6\left(\frac{5 - 4}{20}\right)\\ &=\frac{6}{20} \end{aligned}$$

Variance

$$\begin{aligned} Var(X)&=E(X^2)-(E(X))^2 \\ &=\frac{6}{20}-\left(\frac{1}{2}\right)^2\\ &=0.3-0.25\\ &=0.05 \end{aligned}$$