Question

For the random variable Xwith the following probability density function ⎩⎨⎧<≥=−0x;00x;e2)x(fx2, find i) )1|X|(P>μ− ii) Use Chebyshev’s inequality to obtain an upper bound on ]1|X|[P>μ− and compare with the result in (i).

Answer 1

Answer on Question #65108 – Math – Statistics and Probability

Question

For the random variable $X$ with the following probability density function

$$f(x) = \begin{cases} 0, & x < 0, \\ \frac{x}{2}, & 0 \leq x \leq 2, \\ 0, & x > 2. \end{cases}$$

i) Find $P(|X - E(X)| > 2)$;

ii) Use Chebyshev's inequality to obtain an upper bound on $P(|X - E(X)| > 2)$ and compare with the result in i).

Solution

i) By formula (45) on p. 235 in [1]

$$E(X) = \int_{-\infty}^{+\infty} x f(x) \, dx = \int_{0}^{2} x \cdot \frac{x}{2} \, dx = \left. \frac{x^3}{6} \right|_{0}^{2} = \frac{4}{3}.$$

Then

$$\begin{aligned} P(|X - E(X)| > 2) &= P\left(\left|X - \frac{4}{3}\right| > 2\right) \\ &= P\left(-\infty < X - \frac{4}{3} < -2\right) + P\left(2 < X - \frac{4}{3} < +\infty\right) = \\ &\quad P\left(-\infty < X < -\frac{2}{3}\right) + P\left(\frac{10}{3} < X < +\infty\right) = \\ &= \int_{-\infty}^{-2/3} f(x) \, dx + \int_{10/3}^{+\infty} f(x) \, dx = 0. \end{aligned}$$

ii) We use Chebyshev's inequality ([1], eq. (23) on p. 229):

$$P(|X - E(X)| > 2) \leq \frac{E(X - E(X))^2}{2^2}.$$

We have

$$E(X - E(X))^2 = E(X)^2 - (E(X))^2.$$

Applying formula (45) on p. 235 in [1] again, we get

$$\begin{aligned} E(X - E(X))^2 &= E(X)^2 - (E(X))^2 = \int_{0}^{2} x^2 \cdot \frac{x}{2} \, dx - \left(\frac{4}{3}\right)^2 = \left. \frac{x^4}{8} \right|_{0}^{2} - \frac{16}{9} = 2 - \frac{16}{9} \\ &= \frac{2}{9}. \end{aligned}$$

Then

$$P(|X - E(X)| > 2) \leq \frac{2}{9 \cdot 4} = \frac{1}{18} = 0.05555556.$$

Answer: i) 0, ii) 0.05555556. The probability estimated with Chebyshev's inequality in ii) is small but it does not equal 0 as in i).

References:

[1]. A. N. Shiryaev. Probability-1. Third edition. Springer, 2016.

Answer provided by https://www.AssignmentExpert.com