Question 7.4: Determine in an efficient manner the fourth moment of the ra......
Determine in an efficient manner the fourth moment of the random variable X whose PDF is given by
\,\displaystyle f_X\left (x\right )=32 x^2 e^{-4 x} \quad x \geq 0The easiest method of solving this problem is via the transform method. Thus, we first need to obtain the s-transform of the PDF. We shall use four methods to do this. The first method is a brute-force method, and the other three are “smart” methods.
a. Brute-Force Method In this method
we attempt to obtain MX(s) directly as follows:
Let \,\displaystyle u=x^2 \Rightarrow d u=2 x d x and let \,\displaystyle d v=e^{-\left (s+4\right ) x} d x \Rightarrow v=-e^{-\left (s+4\right ) x} /\left (s+4\right ). Thus,
\,\displaystyle M_X\left (s\right )=32\left[-\frac{x^2 e^{-\left (s+4\right ) x}}{s+4}\right]_0^{\infty}+\frac{32}{s+4} \int_0^{\infty} 2 x e^{-\left (s+4\right ) x} d x=\frac{64}{s+4} \int_0^{\infty} x e^{-\left (s+4\right ) x} d x\\
Let \,\displaystyle u=x \Rightarrow d u=d x, and let \,\displaystyle d v=e^{-\left (s+4\right ) x} d x \Rightarrow v=-e^{-\left (s+4\right ) x} /\left (s+4\right ). Thus,
\,\displaystyle \begin{aligned} M_X\left (s\right ) & =\frac{64}{s+4}\left\{\left[-\frac{x e^{-\left (s+4\right ) x}}{s+4}\right]_0^{\infty}+\int_0^{\infty} \frac{e^{-\left (s+4\right ) x}}{s+4} d x\right\}=\frac{64}{\left (s+4\right )^2} \int_0^{\infty} e^{-\left (s+4\right ) x} d x \\ & =\frac{64}{\left (s+4\right )^2}\left[-\frac{e^{-\left (s+4\right ) x}}{s+4}\right]_0^{\infty}=\frac{64}{\left (s+4\right )^3}=\frac{4^3}{\left (s+4\right )^3}=\left (\frac{4}{s+4}\right )^3 \end{aligned}
b. Smart Method 1
In this method we exploit the properties of the moments of the exponential distribution in carrying out the integration. Thus, we proceed as follows:
Let \mu =s+4 . Then we have that
\,\displaystyle M_X\left (s\right )=32 \int_0^{\infty} x^2 e^{-\mu x} d x=\frac{32}{\mu} \int_0^{\infty} \mu x^2 e^{-\mu x} d x
Now, the integration term is essentially the second moment of an exponential random variable X with parameter \mu. From the results in Chapter 4 we know that for an exponential random variable,
\,\displaystyle E\left[X^n\right]=\frac{n !}{\mu^n} \quad n=1,2, \ldots(See Equation 4.33.) Thus we have that
\,\displaystyle E\left[X^n\right]=\int_0^{\infty} x^n f_X\left (x\right ) d x=\frac{n !}{\lambda^n} \quad n=1,2, \ldots \qquad (4.33)\\ \,\displaystyle M_X\left (s\right )=\frac{32}{\mu}\left\{\frac{2 !}{\mu^2}\right\}=\frac{64}{\mu^3}=\frac{64}{\left (s+4\right )^3}=\left (\frac{4}{s+4}\right )^3
c. Smart Method 2
In this method we realize that X is an Erlang random variable. But to determine its order and parameter we need to put its PDF in the standard form of the Erlang PDF as follows:
From the power of x we see that k-1=2 , which means that k=3. That is, X is a third-order Erlang random variable. Similarly, from the exponential term we that \lambda =4. Let Y be the underlying exponentially distributed random variable. Then the PDF of Y and its stransform are given by
\,\displaystyle \begin{aligned} & f_Y\left (y\right )=4 e^{-4 y} \quad y \geq 0 \\ & M_Y\left (s\right )=\frac{4}{s+4} \end{aligned}Thus, X=Y_1+Y_2+Y_3 , where the Y_k are independent and identically distributed with the above PDF. Since the s-transform of the PDF of the sum of independent random variables is equal to the product of their s-transforms, we have that
\,\displaystyle M_X\left (s\right )=\left[M_Y\left (s\right )\right]^3=\left (\frac{4}{s+4}\right )^3
d. Smart Method 3
As in the previous methods, we start with
\,\displaystyle M_X\left (s\right )=32 \int_0^{\infty} x^2 e^{-\left (s+4\right ) x} d x=32 \int_0^{\infty} x^2 e^{-\mu x} d xBecause the integrand (or integral kernel) looks like an Erlang-3 PDF, we rearrange it to obtain a true Erlang-3 PDF as follows:
\,\displaystyle x^2 e^{-\mu x}=\left\{\frac{\mu^3 x^2 e^{-\mu x}}{2 !}\right\}\left (\frac{2 !}{\mu^3}\right )=\frac{2}{\mu^3}\left\{\frac{\mu^3 x^2 e^{-\mu x}}{2 !}\right\}=\frac{2}{\mu^3} f_{X_3}\left (x\right )\\
This means that
\,\displaystyle \begin{aligned} M_X\left (s\right ) & =32 \int_0^{\infty} x^2 e^{-\mu x} d x=\frac{\left (32\right )\left (2\right )}{\mu^3} \int_0^{\infty} \frac{\mu^3 x^2}{2 !} e^{-\mu x} d x=\frac{64}{\mu^3} \int_0^{\infty} \frac{\mu^3 x^2}{2 !} e^{-\mu x} d x \\ & =\frac{64}{\mu^3}=\frac{4^3}{\mu^3}=\left (\frac{4}{\mu}\right )^3=\left (\frac{4}{s+4}\right )^3 \end{aligned}
where the first equality of the second line follows from the fact that the integral is equal to 1 because a valid PDF is integrated over the entire range of the values of x . The Rest of the Solution: Having obtained the s-transform of the PDF, the fourth moment of X is given by
\,\displaystyle \begin{aligned} E\left[X^4\right] & =\left.\left (-1\right )^4 \frac{d^4}{d s^4} M_X\left (s\right )\right\vert _{s=0}=\left.\frac{d^4}{d s^4}\left (\frac{4}{s+4}\right )^3\right\vert _{s=0} \\ & =\left.\frac{\left (-3\right )\left (-4\right )\left (-5\right )\left (-6\right )\left (4\right )^3}{\left (s+4\right )^7}\right\vert _{s=0}=\left.\frac{\left (360\right )\left (4\right )^3}{\left (s+4\right )^7}\right\vert _{s=0}=1.40625 \end{aligned}