$f_X(x) = \begin{cases} \frac{e}{2(e - 1)}e^{-x}, & 0 < x < 1, \\ \frac{1}{2}, & 1 < x < 2, \\ 0, & \text{elsewhere}. \end{cases}$
Then, the value of the expectation of $X, E[X]$ (rounded off up to two decimal places) is __________
Calculating Expectation E[X] for Piecewise PDF
The expectation of a continuous random variable \(X\) is defined as \( E[X] = \int_{-\infty}^{\infty} x f_X(x) dx \). Given the probability density function (pdf):
$ f_X(x) = \begin{cases} \frac{e}{2(e - 1)}e^{-x}, & 0 < x < 1, \\ \frac{1}{2}, & 1 < x < 2, \\ 0, & \text{elsewhere}. \end{cases} $
We compute the expectation by integrating over the defined intervals where the pdf is non-zero.
Step 1: Integrate Over \( 0 < x < 1 \)
The contribution from the first interval ($0 < x < 1$) is:
$ I_1 = \int_{0}^{1} x \cdot \frac{e}{2(e - 1)}e^{-x} dx $
Factor out the constant:
$ I_1 = \frac{e}{2(e - 1)} \int_{0}^{1} x e^{-x} dx $
Using integration by parts (\( \int u \, dv = uv - \int v \, du \)) with \( u=x \) and \( dv=e^{-x}dx \), we get \( du=dx \) and \( v=-e^{-x} \). The integral of \( x e^{-x} \) is \( -x e^{-x} - e^{-x} \).
Evaluate the definite integral:
$ \int_{0}^{1} x e^{-x} dx = [-x e^{-x} - e^{-x}]_{0}^{1} = \left[ -(x+1)e^{-x} \right]_{0}^{1} $
$ = \left( -(1+1)e^{-1} \right) - \left( -(0+1)e^{0} \right) = -2e^{-1} - (-1) = 1 - \frac{2}{e} $
Substitute this back into \( I_1 \):
$ I_1 = \frac{e}{2(e - 1)} \left( 1 - \frac{2}{e} \right) = \frac{e}{2(e - 1)} \left( \frac{e - 2}{e} \right) = \frac{e - 2}{2(e - 1)} $
Step 2: Integrate Over \( 1 < x < 2 \)
The contribution from the second interval ($1 < x < 2$) is:
$ I_2 = \int_{1}^{2} x \cdot \frac{1}{2} dx $
$ I_2 = \frac{1}{2} \left[ \frac{x^2}{2} \right]_{1}^{2} $
$ = \frac{1}{2} \left( \frac{2^2}{2} - \frac{1^2}{2} \right) = \frac{1}{2} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{1}{2} \left( \frac{3}{2} \right) = \frac{3}{4} $
Step 3: Calculate Total Expectation
The total expectation is the sum of \( I_1 \) and \( I_2 \):
$ E[X] = I_1 + I_2 = \frac{e - 2}{2(e - 1)} + \frac{3}{4} $
Using the approximation \( e \approx 2.718 \):
$ E[X] \approx \frac{2.718 - 2}{2(2.718 - 1)} + \frac{3}{4} $
$ E[X] \approx \frac{0.718}{2(1.718)} + 0.75 \approx \frac{0.718}{3.436} + 0.75 $
$ E[X] \approx 0.20896 + 0.75 \approx 0.95896 $
Rounding to two decimal places, \( E[X] \approx 0.96 \). This value lies between 0.9 and 1.1.
Important Questions from Continuous Distributions
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
A nationalized bank has found that the daily balance available in its savings accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders, who maintain an average daily balance more than Rs 500 is _______
The number of parameters in the univariate exponential and Gaussian distributions, respectively are
Find the value of λ such that the function f (x) is a valid probability density function. _______
\(f\left( x \right)\begin{array}{*{20}{c}} { = \lambda \left( {x - 1} \right)\left( {2 - x} \right)}&{for1 \le x \le 2}\\ { = 0}&{otherwise} \end{array}\)
