Question

The density function of a continuous random variable X is given by

Ae^(-x) , 0≤x≤1
f(x)= 0 , otherwise


Evaluate the following:
a) Mean
b) Variance
c) E [(2+3x)^2]

Answer 1

Answer on Question #51468 - Math - Statistics and Probability

The density function of a continuous random variable $X$ is given by

$$f(x) = \begin{cases} Ae^{-x}, & 0 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases}$$

Evaluate the following:

a) Mean

b) Variance

c) $E[(2 + 3x)^2]$

Solution

Since $f(x)$ is a density function, it must satisfy the following relation

$$\int_{0}^{1} f(x) \, dx = 1$$

Therefore

$$\int_{0}^{1} Ae^{-x} \, dx = A(1 - e^{-1}) = 1 \Rightarrow A = \frac{1}{1 - e^{-1}} = \frac{e}{e - 1}$$

a) Mean

$$\begin{aligned} E[X] &= \mu = \int_{0}^{1} x f(x) \, dx = \int_{0}^{1} x Ae^{-x} \, dx = -A \int_{0}^{1} x \, de^{-x} = -A \left(x e^{-x} \Big|_{0}^{1} - \int_{0}^{1} e^{-x} \, dx\right) \\ &= -A \left(x e^{-x} \Big|_{0}^{1} + e^{-x} \Big|_{0}^{1}\right) = \\ &= -A \left(e^{-1} - 0 + e^{-1} - 1\right) = A \left(1 - 2e^{-1}\right) = \frac{1 - 2e^{-1}}{1 - e^{-1}} = \frac{e - 2}{e - 1} \end{aligned}$$

b) Variance

$$Var[X] = \sigma^{2} = \int_{0}^{1} (x - \mu)^{2} f(x) \, dx = \int_{0}^{1} Ae^{-x} (x - \mu)^{2} \, dx = \int_{0}^{1} Ae^{-x} \left(x^{2} - 2x\mu + \mu^{2}\right) \, dx =$$ $$\begin{array}{l} = \int_{0}^{1} A e^{-x} x^{2} dx - \int_{0}^{1} A e^{-x} 2x \mu dx + \int_{0}^{1} A e^{-x} \mu^{2} dx = A \int_{0}^{1} e^{-x} x^{2} dx - 2\mu \int_{0}^{1} A e^{-x} x dx \\ \quad + \mu^{2} \int_{0}^{1} A e^{-x} dx = \int_{0}^{1} A e^{-x} x^{2} dx - 2\mu^{2} + \mu^{2} = \int_{0}^{1} A e^{-x} x^{2} dx - \mu^{2} \\ = E[X^{2}] - \mu^{2} \end{array}$$ $$\begin{array}{l} E[X^{2}] = \int_{0}^{1} A e^{-x} x^{2} dx = -A \int_{0}^{1} x^{2} d e^{-x} = -A x^{2} e^{-x} \Big|_{0}^{1} + 2A \int_{0}^{1} x e^{-x} dx = \\ \quad = -A e^{-1} + 2A(1 - 2e^{-1}) = A(2 - 5e^{-1}) \end{array}$$ $$\begin{array}{l} \int_{0}^{1} A e^{-x} 2x \mu dx = 2\mu \left(A \int_{0}^{1} x e^{-x} dx\right) = 2\mu \cdot \mu = 2\mu^{2} \\ \int_{0}^{1} A e^{-x} \mu^{2} dx = \mu^{2} \int_{0}^{1} A e^{-x} dx = \mu^{2} \int_{0}^{1} f(x) dx = \mu^{2} \\ \sigma^{2} = A(2 - 5e^{-1}) - 2\mu^{2} + \mu^{2} = \\ = \frac{2 - 5e^{-1}}{1 - e^{-1}} - \mu^{2} = \frac{2 - 5e^{-1}}{1 - e^{-1}} - \left(\frac{1 - 2e^{-1}}{1 - e^{-1}}\right)^{2} = \frac{1 - 3e^{-1} + 4e^{-2}}{(1 - e^{-1})^{2}} \end{array}$$

c) $E[(2 + 3X)^{2}]$

$$\begin{array}{l} E[(2 + 3X)^{2}] = E[4 + 12X + 9X^{2}] = E[4] + E[12X] + E[9X^{2}] = 4 + 12E[X] + 9E[X^{2}] \\ \quad = 4 + 12\mu + 9(\text{Var}[X] + \mu^{2}) = 4 + 12\frac{e - 2}{e - 1} + 9A(2 - 5e^{-1}) \\ \quad = 4 + 12\frac{e - 2}{e - 1} + 9\frac{e}{e - 1}\frac{2e - 5}{e} = \frac{4e - 4 + 12e - 24 + 18e - 45}{e - 1} \\ = \frac{34e - 73}{e - 1} = \frac{34 - 73e^{-1}}{1 - e^{-1}} \end{array}$$

because

$$\begin{array}{l} E[(2 + 3X)^{2}] = \int_{0}^{1} A e^{-x} (3x + 2)^{2} dx = \int_{0}^{1} A e^{-x} 9x^{2} dx + \int_{0}^{1} A e^{-x} 12x dx + \int_{0}^{1} A e^{-x} 4dx = \\ \quad = 9A(2 - 5e^{-1}) + 12A(1 - 2e^{-1}) + 4A(1 - e^{-1}) = \frac{34 - 73e^{-1}}{1 - e^{-1}} \end{array}$$

**Answer**

a) $\mu = \frac{1 - 2e^{-1}}{1 - e^{-1}}$

b) $\sigma^{2} = \frac{1 - 3e^{-1} + 4e^{-2}}{(1 - e^{-1})^{2}}$

c)

$$E[(2 + 3x)^2] = \frac{34 - 73e^{-1}}{1 - e^{-1}}$$

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