Question

For the random variable X with the following probability density function
f (x) ={2e^(-2x); 0 is greater than equal to x and 0; 0 is smaller than x
find
i) ) P (| X − μ | >1
ii) Use Chebyshev’s inequality to obtain an upper bound on P[| X − μ | >1] and
compare with the result in (i).

Answer 1

Answer on Question #65598 - Math - Statistics and Probability

**Question**: For the random variable $X$ with the following probability density function

$$f(x) = \begin{cases} 2e^{-2x}, & 0 \geq x; \\ 0, & 0 < x \end{cases}$$

find

i) $P\{|X - \mu| > 1\}$;

ii) Use Chebyshev's inequality to obtain an upper bound on $P\{|X - \mu| > 1\}$ and compare with the result in (i).

**Solution**: First of all the function $f$ is not probability density function for any random variable $X$. Indeed,

$$\int_{-\infty}^{+\infty} f(x) \, dx = \int_{-\infty}^{0} 2e^{-2x} \, dx = \int_{-\infty}^{0} e^{-2x} \, d(2x) = -e^{-2x} \Big|_{-\infty}^{0} = +\infty \neq 1.$$

Therefore, the correct probability density function (pdf) should be defined as follows:

$$f(x) = \begin{cases} 2e^{-2x}, & x \geq 0; \\ 0, & x < 0 \end{cases}$$

or

$$f(x) = \begin{cases} 2e^{2x}, & 0 \geq x; \\ 0, & 0 < x. \end{cases}$$

I will consider the first function

$$f(x) = \begin{cases} 2e^{-2x}, & x \geq 0; \\ 0, & x < 0, \end{cases}$$

which is a probability density function of the exponential distribution.

The expectation for $X$ is equal to

$$\begin{aligned} \mu &= \int_{-\infty}^{+\infty} x f(x) \, dx \\ &= \int_{0}^{+\infty} x \cdot 2e^{-2x} \, dx = - \int_{0}^{+\infty} x \, d(e^{-2x}) = - \left( x e^{-2x} \Big|_{0}^{+\infty} - \int_{0}^{+\infty} e^{-2x} \, dx \right) \\ &= - \left( \lim_{x \to +\infty} \frac{x}{e^{2x}} - 0 + \frac{1}{2} e^{-2x} \Big|_{0}^{+\infty} \right) = 0 - \frac{1}{2} \left( \lim_{x \to +\infty} e^{-2x} - 1 \right) = - \frac{1}{2} (0 - 1) = \frac{1}{2}. \end{aligned}$$ $$\begin{aligned} P\{|X - \mu| > 1\} &= P(\{X > \mu + 1\} \cup \{X < \mu - 1\}) \\ &= P\left(\{X > \frac{3}{2}\} \cup \{X < -\frac{1}{2}\}\right) \\ \int_{3/2}^{+\infty} 2e^{-2x} \, dx + \int_{-\infty}^{-1/2} 0 \, dx &= \int_{3/2}^{+\infty} e^{-2x} \, d(2x) \\ &= -e^{-2x} \Big|_{\frac{3}{2}}^{+\infty} = e^{-2 \cdot \frac{3}{2}} = e^{-3}. \end{aligned}$$

ii) By Chebyshev's inequality we have $P\{|X - \mu| > 1\} \leq \frac{Var[X]}{3^2} = Var[X]$.

Let us calculate the variance $Var[X] = E[X^2] - (E[X])^2$.

$$\begin{aligned} E[X^2] &= \int_{-\infty}^{+\infty} x^2 \cdot f(x) \, dx = \int_{0}^{+\infty} x^2 \cdot 2e^{-2x} \, dx = - \int_{0}^{+\infty} x^2 \, d(e^{-2x}) \\ &= - \left( x^2 e^{-2x} \Big|_{0}^{+\infty} - \int_{0}^{+\infty} e^{-2x} \cdot 2x \, dx \right) = \frac{1}{2} = \frac{1}{2} \cdot 2x. \end{aligned}$$ $$Var[X] = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{4}.$$

Hence, $P\{|X - \mu| > 1\} \leq 0.25$ by Chebyshev's inequality.

Finally, we have that the exact value of probability $P\{|X - \mu| > 1\}$ equals $e^{-3} \approx 0.049787$ and the approximation obtained by Chebyshev's inequality is equal to 0.25. This gives

$$\frac{|0.049787 - 0.25|}{0.049787} \cdot 100\% \approx 402\%$$

of percent error. In other words, the upper bound is 5 times greater than the exact value of probability.

Answer: $P\{|X - \mu| > 1\} = e^{-3} \approx 0.049787$, $P\{|X - \mu| > 1\} \leq 0.25$ by Chebyshev's inequality.

For Chebyshev's inequality see for example

W. Feller, "An introduction to probability theory and its applications", 1, Wiley (1957-1971), P.233, and for pdf and its properties

W. Feller, "An introduction to probability theory and its applications", 2, Wiley (1971), P.3-5

https://www.encyclopediaofmath.org/index.php/Feller_%22An_introduction_to_probability_theory_and_its_applications%22

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