Characteristic function

Question

The characteristic function for the random variable X is $ f (x) = cos ^ 3 (x) $. Find a characteristic function for the random variable $ Y = X ^ 3-3X ^ 2 + 1 $ .. Can I have instructions on how this task is done?

Answer 1

In the first step, we must determine the domain of $X$. I suppose $X\geq 0$. Thus, we have:

$\int_{0}^{b}{f(x)dx} = 1 \rightarrow \int_{0}^{b}{cos^{3}(x)dx} = 1 \rightarrow \frac{1}{12}\big(9sin(x)+sin(3x)\big)\Big]_{0}^{b} = 1$

By the above computation, we can compute the value of $b$. Then, we have

$Y = X^{3} - 3 X^{2} + 1 \rightarrow itY = itX^{3} - 3it X^{2} + 1 \rightarrow e^{itY} = e^{itX^{3} - 3it X^{2} + it} \rightarrow E[e^{itY}] = E[e^{itX^{3} - 3it X^{2} + it}] = \int_{0}^{b}{ e^{itx^{3} - 3it x^{2} + it}f_{X}(x)dx } = \int_{0}^{b}{ e^{itx^{3} - 3it x^{2} + it}cos^{3}(x)dx }$

By solving the above integral, the characteristic function of $Y$ obtained.

Answer 2

The charateristic function can be written as \begin{align} f(x)&=\cos^3(x)\\ &=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^3\\ &=\frac18e^{3ix}+\frac38e^{ix}+\frac38e^{-ix}+\frac18e^{-3ix} \end{align} This indicates that your random variable is a discrete random variable which take the values of $-3,-1,1,3$ with probabilities $\frac18, \frac38,\frac38$ and $\frac18$, repectively.

From these you can easily calulate the char function of $Y=X^3-3x^2+1$ to be \begin{align} Ee^{iY\theta}&=\frac18\times e^{i\theta\left((-3)^3-3(-3)^2+1\right)}+\frac38\times e^{i\theta\left((-1)^3-3(-1)^2+1\right)}\\&+\frac38\times e^{i\theta\left((1)^3-3(1)^2+1\right)}+\frac18\times e^{i\theta\left((3)^3-3(3)^2+1\right)} \end{align}