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4.1.4 Solved Problems:
Continuous Random Variables

Problem

Let $X$ be a random variable with PDF given by \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} cx^2& \quad |x| \leq 1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}

  1. Find the constant $c$.
  2. Find $EX$ and Var$(X)$.
  3. Find $P(X \geq \frac{1}{2})$.

  • Solution

      1. To find $c$, we can use $\int_{-\infty}^{\infty} f_X(u)du=1$:
        $1$ $=\int_{-\infty}^{\infty} f_X(u)du$
        $= \int_{-1}^{1} cu^2du$
        $= \frac{2}{3} c.$

        Thus, we must have $c=\frac{3}{2}$.
      2. To find $EX$, we can write
        $EX$ $= \int_{-1}^{1} u f_X(u)du$
        $= \frac{3}{2}\int_{-1}^{1} u^3 du$
        $=0.$

        In fact, we could have guessed $EX=0$ because the PDF is symmetric around $x=0$. To find Var$(X)$, we have
        $\textrm{Var}(X)$ $=EX^2-(EX)^2=EX^2$
        $= \int_{-1}^{1} u^2 f_X(u)du$
        $= \frac{3}{2}\int_{-1}^{1} u^4 du$
        $=\frac{3}{5}.$

      3. To find $P(X \geq \frac{1}{2})$, we can write $$P(X \geq \frac{1}{2})=\frac{3}{2} \int_{\frac{1}{2}}^{1} x^2dx=\frac{7}{16}.$$


Problem

Let $X$ be a continuous random variable with PDF given by $$f_X(x)=\frac{1}{2}e^{-|x|}, \hspace{20pt} \textrm{for all }x \in \mathbb{R}.$$ If $Y=X^2$, find the CDF of $Y$.

  • Solution

    • First, we note that $R_Y=[0,\infty)$. For $y \in [0,\infty)$, we have

      $F_Y(y)$ $=P(Y \leq y)$
      $=P(X^2 \leq y)$
      $=P(-\sqrt{y} \leq X \leq \sqrt{y})$
      $=\int_{-\sqrt{y}}^{\sqrt{y}} \frac{1}{2}e^{-|x|} dx$
      $=\int_{0}^{\sqrt{y}} e^{-x} dx$
      $=1-e^{-\sqrt{y}}.$

      Thus, \begin{equation} \nonumber F_Y(y) = \left\{ \begin{array}{l l} 1-e^{-\sqrt{y}} & \quad y \geq 0\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}


Problem

Let $X$ be a continuous random variable with PDF \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} 4x^3 & \quad 0 < x \leq 1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Find $P(X \leq \frac{2}{3} | X> \frac{1}{3})$.

  • Solution

    • We have

      $P(X \leq \frac{2}{3} | X > \frac{1}{3})$ $=\frac{P(\frac{1}{3} < X \leq \frac{2}{3})}{P(X > \frac{1}{3})}$
      $=\frac{\int_{\frac{1}{3}}^{\frac{2}{3}} 4x^3 dx}{\int_{\frac{1}{3}}^{1} 4x^3 dx}$
      $=\frac{3}{16}.$


Problem

Let $X$ be a continuous random variable with PDF \begin{equation} \nonumber f_X(x) = \left\{ \begin{array}{l l} x^2\left(2x+\frac{3}{2}\right) & \quad 0 < x \leq 1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} If $Y=\frac{2}{X}+3$, find Var$(Y)$.

  • Solution

    • First, note that $$\textrm{Var}(Y)=\textrm{Var}\left(\frac{2}{X}+3\right)=4\textrm{Var}\left(\frac{1}{X}\right), \hspace{15pt} \textrm{using Equation 4.4}$$ Thus, it suffices to find Var$(\frac{1}{X})=E[\frac{1}{X^2}]-(E[\frac{1}{X}])^2$. Using LOTUS, we have $$E\left[\frac{1}{X}\right]=\int_{0}^{1} x\left(2x+\frac{3}{2}\right) dx =\frac{17}{12}$$ $$E\left[\frac{1}{X^2}\right]=\int_{0}^{1} \left(2x+\frac{3}{2}\right) dx =\frac{5}{2}.$$ Thus, Var$\left(\frac{1}{X}\right)=E[\frac{1}{X^2}]-(E[\frac{1}{X}])^2=\frac{71}{144}$. So, we obtain $$\textrm{Var}(Y)=4\textrm{Var}\left(\frac{1}{X}\right)=\frac{71}{36}.$$


Problem

Let $X$ be a positive continuous random variable. Prove that $EX=\int_{0}^{\infty} P(X \geq x) dx$.

  • Solution

    • We have $$P(X \geq x)=\int_{x}^{\infty}f_X(t)dt.$$ Thus, we need to show that $$\int_{0}^{\infty} \int_{x}^{\infty}f_X(t)dtdx=EX.$$ The left hand side is a double integral. In particular, it is the integral of $f_X(t)$ over the shaded region in Figure 4.4.

      Fig.4.4 - The shaded area shows the region of the double integral of Problem 5.
      We can take the integral with respect to $x$ or $t$. Thus, we can write
      $\int_{0}^{\infty} \int_{x}^{\infty}f_X(t)dtdx$ $=\int_{0}^{\infty} \int_{0}^{t}f_X(t)dx dt$
      $=\int_{0}^{\infty} f_X(t) \left(\int_{0}^{t} 1 dx \right) dt$
      $=\int_{0}^{\infty} tf_X(t) dt=EX \hspace{20pt} \textrm{since $X$ is a positive random variable}.$


Problem

Let $X \sim Uniform(-\frac{\pi}{2},\pi)$ and $Y=\sin(X)$. Find $f_Y(y)$.

  • Solution

    • Here $Y=g(X)$, where $g$ is a differentiable function. Although $g$ is not monotone, it can be divided to a finite number of regions in which it is monotone. Thus, we can use Equation 4.6. We note that since $R_X=[-\frac{\pi}{2},\pi]$, $R_Y=[-1,1]$. By looking at the plot of $g(x)=\sin(x)$ over $[-\frac{\pi}{2},\pi]$, we notice that for $y \in (0,1)$ there are two solutions to $y=g(x)$, while for $y \in (-1,0)$, there is only one solution. In particular, if $y \in (0,1)$, we have two solutions: $x_1=\arcsin(y)$, and $x_2=\pi-\arcsin(y)$. If $y \in (-1,0)$ we have one solution, $x_1=\arcsin(y)$. Thus, for $y \in(-1,0)$, we have

      $f_Y(y)$ $= \frac{f_X(x_1)}{|g'(x_1)|}$
      $= \frac{f_X(\arcsin(y))}{|\cos(\arcsin(y))|}$
      $= \frac{\frac{2}{3 \pi}}{\sqrt{1-y^2}}.$

      For $y \in(0,1)$, we have
      $f_Y(y)$ $= \frac{f_X(x_1)}{|g'(x_1)|}+\frac{f_X(x_2)}{|g'(x_2)|}$
      $= \frac{f_X(\arcsin(y))}{|\cos(\arcsin(y))|}+\frac{f_X(\pi-\arcsin(y))}{|\cos(\pi-\arcsin(y))|}$
      $= \frac{\frac{2}{3 \pi}}{\sqrt{1-y^2}}+\frac{\frac{2}{3 \pi}}{\sqrt{1-y^2}}$
      $= \frac{4}{3 \pi \sqrt{1-y^2}}.$

      To summarize, we can write \begin{equation} \nonumber f_Y(y) = \left\{ \begin{array}{l l} \frac{2}{3 \pi \sqrt{1-y^2}} & \quad -1 < y < 0\\ \frac{4}{3 \pi \sqrt{1-y^2}} & \quad 0 < y < 1\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}

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