Couple exercises from Functions of Random Variables
Created on July 17, 2023
Last modified on July 18, 2023
Written by Some author
Read time: 6 minutes
Summary: Examples include finding the probability density functions of transformed random variables, absolute values, sums of independent variables, and squared variables, as well as obtaining the cumulative distribution function, applying the moment generating function theorem, and determining the distributions of sums of Poisson and exponential random variables.
$\textbf{Example}$. If the pdf of $X$ is
$$ f(x)= \begin{cases}6 x(1-x), & \text { if } 0<x<1 \\ 0, & \text { otherwise }\end{cases} $$
Find the probability density function of $Y=X^3$.
Then find the probability density function of $Y=X^3$.
$$\Pr[Y \le k] = \Pr[X^3 \le k] = \Pr[X \le k^{1/3}]$$
We know that $\int_{0}^{k^{1/3}} 6x(1-x) = \int_{0}^{k^{1/3}} 6x-6x^2 = 3(k^{1/3})^2 -2(k^{1/3})^3 = 3 k^{2/3} - 2k $
So the pdf of $Y$ will be $f(k) = 2k^{-1/3}-2$, $k \in(0,1)$
$\textbf{Example.}$ Let $X$ be a random variable with pdf $f(x)$ and $Y=|X|$. Show that the pdf of $Y$ is
$$ g(y)= \begin{cases}f(y)+f(-y), & \text { if } y>0 \\ 0, & \text { otherwise }\end{cases} $$
Let's consider $\Pr[Y \le k] = \Pr[|X| \le k] = \Pr[-k \le X \le k] = \int_{-k}^k f(y)\, dy$
Then if we take derivative with respect to $k$, we will have $$\frac{d}{dk}\Pr[Y \le k] = \frac{d}{dk}\int_{-k}^k f(y)\, dy = f(k) -(-f(-k)) = f(k) +f(-k)$$
$\textbf{Example.}$ Use the previous result to find the pdf of $Y=|X|$ where $X$ is the standard normal RV.
$$X: f(x) = \frac{1}{\sqrt{2 \pi}} \exp(-\frac{x^2}{2})$$
So $$Y: f(y) = \frac{\sqrt{2}}{\sqrt{\pi}} \exp(-\frac{y^2}{2})$$
$\textbf{Example.}$ Suppose the joint pdf of $\left(X_1, X_2\right)$ is
$$f\left(x_1, x_2\right)= \begin{cases}6 e^{-3 x_1-2 x_2}, & \text { if } x_1, x_2>0 \\ 0, & \text { otherwise. }\end{cases}$$
Find the pdf of $Y=X_1+X_2$.
Let's consider $$\Pr[Y \le k] = \Pr[X_1 +X_2 \le k] = \int_0^{k} \int_{0}^{k-y} 6 \exp(-3x -2y)\, dx \,dy$$
$$= 6\int_0^k \exp(-2y) \int_{0}^{k-y}\exp(-3x)\, dx \, dy$$
$$ = -2\int_0^k \exp(-2y) \exp(-3x) |_{0}^{k-y} \, dy$$
$$= -2\int_0^k \exp(-2y) (\exp(-3(k-y)) - 1) \, dy$$
$$=-2 \int_0^k (\exp(-3k+y)) - \exp(-2y)) \, dy$$
$$= -2 (\exp(-3k + y) + \frac{1}{2}\exp(-2y) |_0^k)$$
$$= -2 (\exp(-2k ) + \frac{1}{2}\exp(-2k) - \exp(-3k ) - \frac{1}{2} )$$
$$= -3\exp(-2k) +2\exp(-3k) +1$$
And if we take derivative with respect to $k$, we will have $f(k) = 6 \exp(-2k) -6\exp(-3k)$ for $k > 0$.
Now, let's look into some discrete distribution.
$\textbf{Example.}$ Let $X$ be the number of heads by tossing a fair coin for 4 times. Then the pmf $f$ of $X$ is
x,f(x)
0,0.0625
1,0.25
2,0.375
3,0.25
4,0.0625
Find the pmf $g$ of $Y=\frac{1}{1+X}$.
So, we know that function $\frac{1}{1+X}$ has unique value over $[0, \infty)$,
so we will have
$$y = 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$$
and
$$f(y) = \frac{1}{16}, \frac{1}{4}, \frac{3}{8}, \frac{1}{4}, \frac{1}{16}$$
$\textbf{Example.}$ Now, let's compute $Z = (X-2)^2.$
$z = 0, 1, 4$
$f_{Z}(z) = \frac{3}{8}, \frac{1}{2}, \frac{1}{8}$
Theorem. Let $X$ be continuous with pdf $f(x)$ and $Y=u(X)$ where $u$ is strictly monotone in $\operatorname{supp}(f)$, then pdf $g(y)$ of $Y$ is
$$g(y)=f(w(y))\left|w^{\prime}(y)\right|$$
where $w$ is the inverse of $u$ (i.e., $y=u(x)$ iff $x=w(y)$ ).
Proof.
Assume that $u$ is monotonically increasing. and we are dealing with positive support.
$$\Pr[Y \le y] = \Pr[u(X) \le y] = \Pr[X \le w(y)]$$
If we take derivative both sides with respect to $y$, we will have
$$g(y) = f'(\omega(y)) |\omega'(y)|$$
In the same way, we can prove that for non-increasing function, we will have the same result.
$\textbf{ Example }$. Let $X$ be the standard exponent RV. Find the pdf of $Y = \sqrt{X}$.
Let's consider
$$Y = u(X), X = \omega(Y)$$
so we have $$g(x) = 2\lambda xe^{-\lambda x^2} $$
$\textbf{Example.}$ If $F(x)$ is the distribution function of the continuous RV $X$. Find the pdf of $Y=F(X)$.
$u(x) = F(x), $ then $\omega(x) = F^{-1}(x)$
so we know that $F(F^{-1}(x)) = x$, so $$F'(F^{-1}(x)) F'^{-1}(x) =1$$
so $$g(x) = \frac{f(F^{-1}(x))}{F'(F^{-1}(x))}=1$$
Example. Let $X$ be the standard normal random variable. Find the pdf of $Z=X^2$.
$$u(x) = x^2$$, then $\omega(x) = x^{1/2}$ so we will have $\omega'(x) = \frac{1}{2}x^{-1/2}$
so $$g(x) = 2 \times \frac{1}{2}x^{-1/2} \times \frac{1}{\sqrt{2\pi}} \exp(- x/2) = \frac{x^{-1/2}}{\sqrt{2\pi}}\exp(- x/2)$$
$\textbf{Example.}$ Let $\left(X_1, X_2\right)$ have joint density
$$f\left(x_1, x_2\right)= \begin{cases}e^{-\left(x_1+x_2\right)}, & \text { if } x_1, x_2>0, \\ 0, & \text { otherwise. }\end{cases}$$
Find the pdf of $Y=\frac{X_1}{X_1+X_2}$.
Let's consider $$U = \frac{X_1}{X_1+X_2}$$ and $$V = X_1 +X_2$$
then $X_1 = UV$ and $X_2 = V - UV$.
Now the Jacobian matrix will be
$$\begin{pmatrix}V & U\\ -V & 1-U\end{pmatrix}$$,
whose determinant will be $V- UV +UV = V$.
Then we will have $g(u,v)= \exp(-uv - v +uv)v = \exp(-v)v$
Now since we are interested in $u$, we can integrate $\exp(-v)v$ with respect to its support $(0, \infty)$
$$\int v \exp(-v) = -v\exp(-v) +\int \exp(-v) = -v \exp(-v) -\exp(-v)$$
so we will have the probability density function for $Y$ will be $f(y) = 1.$
$\textbf{Example.}$ Let $\left(X_1, X_2\right)$ be uniformly distributed in $(0,1)^2$. Find the pdf of $Y=X_1+X_2$
Let's consider $U = X_1+X_2$, $V = X_2$, then we will have $X_1 = U-V$ and $X_2 = V$.
So the Jacobian matrix will be
$$\begin{pmatrix} 1 & -1 \\ 0 & 1\end{pmatrix}$$
and its determinant is $1$.
So our pdf for $U$ and $V$ is $1$ for $ 0< V < 1$ and $0< U-V < 1$
Now we will integrate with respect to $V$ by fixing our $U$, then we will have
If $u \le 1$:
$$\int_0^u 1 \, dv = u$$
If $u \ge 1$:
$\int_{u-1}^1 1\, dv = 2-u$
Example. Let $\left(X_1, X_2, X_3\right)$ be RVs with joint pdf $f$ as follows:
$$f\left(x_1, x_2, x_3\right)= \begin{cases}e^{-\left(x_1+x_2+x_3\right)}, & \text { if } x_1, x_2, x_3>0 \\ 0, & \text { otherwise }\end{cases}$$
Suppose $Y_1=X_1+X_2+X_3, Y_2=X_2, Y_3=X_3$. Find the marginal pdf of $Y_1$.
$$X_1 = Y_1 - Y_2 - Y_3$$
$$X_2 = Y_2$$
$$X_3 = Y_3$$
$$\begin{pmatrix} 1 & -1 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
So the Jacobian will be $1$.
And we have the pdf for $Y_1, Y_2, Y_3$ will be
$\exp(-y_1)$ with constraints: $ 0 < y_2 < y_1, 0< y_3 < y_1 - y_2$
Now if we integrate with respect to $y_3$, we will get the marginal pdf:
$$\exp(-y_1)(y_1 - y_2) = \exp(-y_1)y_1 - \exp(-y_1)y_2$$
Now if we integrate with respect to $y_2$, we will have the marginal pdf to be
$$g(y_1) = \frac{1}{2}\exp(-y_1)y_1^2 $$
$\textbf{Theorem.}$ The mgf of $Y=X_1+\cdots+X_n$, where $X_1, \ldots, X_n$ are independent RVs with MGFs $M_{X_1}(t), \ldots, M_{X_n}(t)$ respectively, is
$$M_Y(t)=\prod_{i=1}^n M_{X_i}(t)$$.
Proof.
$$\mathbb{E}[\exp(Yt)] =\mathbb{E}[\exp((X_1 + ... +X_n)t)] \\= \mathbb{E}[\exp((X_1t + ... +X_nt)] \\= \mathbb{E}[\prod_{i=1}^n\exp(X_it)] $$
Due to independence, we will have
$$\mathbb{E}[\exp(Yt)] = \prod_{i=1}^n \mathbb{E}[\exp(X_it)] =\prod_{i=1}^n M_{X_i}(t) $$
$\textbf{Example.}$ Suppose $X_1, \ldots, X_n$ are independent Poisson RVs with parameters $\lambda_1, \ldots, \lambda_n$ respectively. Find the distribution of $Y=X_1+\cdots+X_n$.
Let's consider the moment generating function $\mathbb{E}[\exp(Yt)] = \prod_{i=1}^n \exp(\lambda_{i}(e^t-1)) = \exp( (e^t-1)\sum_{i=1}^n \lambda_i) $
So it's a possion distribution with parameter $\sum \lambda_i$
$\textbf{Example.}$ Suppose $X_1, \ldots, X_n$ are independent exponential RVs with the same parameter $\theta$. Find the distribution of $Y=X_1+\cdots+X_n$.
$\mathbb{E}[\exp(Xt)] = \int_0^{\infty} \exp(xt) \frac{1}{\theta}\exp(-\frac{1}{\theta}x) \, dx = \frac{1}{\theta}\int_0^{\infty} \exp(x(t-\frac{1}{\theta})) \, dx = \frac{1}{\theta} \frac{-1}{t-\frac{1}{\theta}} = \frac{1}{1- \theta t} $
So $$\mathbb{E}[\exp(Yt)] = \left(\frac{1}{1-\theta t}\right)^n$$,
which is gamma distribution with $n$,$\theta$
In total: it took me about one hour and 30 minutes to finish this exercises. It's quite slow but I'm getting there slightly.
Other References:
https://people.stat.sc.edu/hitchcock/stat512spring2012.html
https://www.math.wustl.edu/~sawyer/math494s10.html