The skewness coefficient for given pdf?

Question

$$f(x)= \begin{cases} 0.5-\frac{x}{8},& 0\le x \le 4 \\ 0, & \text{otherwise} \end{cases} $$

I have found $E(X) = \frac{4}{3}$ and $Var(X) = \frac{8}{9}$

Th problem says that the skewness coefficiennt is defiend as $E[(X-\mu)^3]/\sigma^3$ and asks to show that its value for the given pdf is 0.566.

It gives a hint:

consider $g(x) = [(X-\mu)^3]/\sigma^3$ and compute the expected value of this $g(x)$

I am completely lost on where to start.

Do I just put my $E(X)$ into $\mu$ and compute?

Side question:

Is $E(x^3) = \int x^3f(x)$ since $E(x) = \int xf(x)$?

Answer 1

You first compute \begin{equation} E[x^3]=\int_{0}^4 0.5x^3-x^4/8 dx=[x^4/8-x^5/40]_{0}^4=32-128/5=32/5=6.4 \end{equation} Now that $E[x^3]$ is known skewness is simply \begin{equation} E[(x-\mu)^3]=E[x^3]-3(\sigma^2+\mu^2)\mu+3\mu^3-\mu^3=6.4-3[\frac{8}{9}+\frac{16}{9}]\frac{4}{3}+2(\frac{4}{3})^3=0.474 \end{equation} Finally normalize this by $\sigma^2=8/9$ to obtain your answer $0.566$.

Answer 2

Yes.   For any deterministic function $g()$ of real-valued continuous random variable $X$ with probability density function $f_X()$, then: $$\mathsf E(g(X)) =\int_\Bbb R g(x)\,f_X(x)\operatorname dx$$

So $$\begin{align} \mu & =\mathsf E(X) \\[1ex] & = \int_\Bbb R x\;f_X(x)\operatorname d x \\[1ex] & = \int_0^4 x(\tfrac 12-\tfrac x 8)\operatorname d x & =\tfrac 4 3 \\[2ex] \sigma^2 & = \mathsf {Var}(X) \\[1ex] & = \int_\Bbb R (x-\mu)^3~f_X(x)\operatorname dx \\[1ex] & =\int_0^4 (x-\mu)^2(\tfrac 12-\tfrac x 8)\operatorname d x & = \tfrac 89 \\[2ex] \mathsf E((X-\mu)^3) & = \int_\Bbb R (x-\mu)^3~f_X(x)\operatorname dx \\[1ex] & = \int_0^4 (x-\mu)^3(\tfrac 12-\tfrac x 8)\operatorname d x\end{align}$$