3.2.5 Solved Problems:
More about Discrete Random Variables
Let $X$ be a discrete random variable with the following PMF \begin{equation} \nonumber P_X(x) = \left\{ \begin{array}{l l} 0.3 & \quad \text{for } x=3\\ 0.2 & \quad \text{for } x=5\\ 0.3 & \quad \text{for } x=8\\ 0.2 & \quad \text{for } x=10\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Find and plot the CDF of $X$.
- Solution
- The CDF is defined by $F_X(x)=P(X \leq x)$. We have \begin{equation} \nonumber F_X(x) = \left\{ \begin{array}{l l} 0 & \quad \text{for } x<3\\ P_X(3)= 0.3 & \quad \text{for } 3 \leq x<5\\ P_X(3)+P_X(5)=0.5 & \quad \text{for } 5 \leq x<8 \\ P_X(3)+P_X(5)+P_X(8)=0.8 & \quad \text{for } 8 \leq x<10 \\ 1 & \quad \text{for } x \geq 10\\ \end{array} \right. \end{equation}
Problem
Let $X$ be a discrete random variable with the following PMF \begin{equation} \nonumber P_X(k) = \left\{ \begin{array}{l l} 0.1 & \quad \text{for } k=0\\ 0.4 & \quad \text{for } k=1\\ 0.3 & \quad \text{for } k=2\\ 0.2 & \quad \text{for } k=3\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
- Find $EX$.
- Find Var$(X)$.
- If $Y=(X-2)^2$, find $EY$.
- Solution
-
-
$EX$ $= \sum_{x_k \in R_X} x_kP_X(x_k)$ $= 0 (0.1)+ 1(0.4)+2(0.3)+3(0.2)$ $=1.6$
- We can use Var$(X)=EX^2-(EX)^2=EX^2-(1.6)^2$. Thus we need to find $EX^2$. Using LOTUS, we have $$EX^2 = 0^2 (0.1)+ 1^2(0.4)+2^2(0.3)+3^2(0.2)=3.4$$ Thus, we have $$\textrm{Var}(X)= (3.4)-(1.6)^2=0.84$$
- Again, using LOTUS, we have $$E(X-2)^2 = (0-2)^2 (0.1)+ (1-2)^2(0.4)+(2-2)^2(0.3)+(3-2)^2(0.2)=1.$$
-
-
Problem
Let $X$ be a discrete random variable with PMF \begin{equation} \nonumber P_X(k) = \left\{ \begin{array}{l l} 0.2 & \quad \text{for } k=0\\ 0.2 & \quad \text{for } k=1\\ 0.3 & \quad \text{for } k=2\\ 0.3 & \quad \text{for } k=3\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} Define $Y=X(X-1)(X-2)$. Find the PMF of $Y$.
- Solution
-
First, note that $R_Y=\{x(x-1)(x-2) | x \in \{0,1,2,3\}\}=\{0,6\}$. Thus,
$P_Y(0)$ $=P(Y=0)=P\big( (X=0) \textrm{ or } (X=1) \textrm{ or } (X=2)\big)$ $=P_X(0)+P_X(1)+P_X(2)$ $=0.7$; $P_Y(6)$ $= P(X=3)=0.3$
Thus, \begin{equation} \nonumber P_Y(k) = \left\{ \begin{array}{l l} 0.7 & \quad \text{for } k=0\\ 0.3 & \quad \text{for } k=6\\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}
-
First, note that $R_Y=\{x(x-1)(x-2) | x \in \{0,1,2,3\}\}=\{0,6\}$. Thus,
Problem
Let $X \sim Geometric(p)$. Find $E\left[\frac{1}{2^X}\right]$.
- Solution
-
The PMF of $X$ is given by
\begin{equation}
\nonumber P_X(k) = \left\{
\begin{array}{l l}
pq^{k-1}& \quad \text{for } k=1,2,3,...\\
0 & \quad \text{otherwise}
\end{array} \right.
\end{equation}
where $q=1-p$. Thus,
$E\left[\frac{1}{2^X}\right]$ $=\sum_{k=1}^{\infty} \frac{1}{2^k} P_X(k)$ $=\sum_{k=1}^{\infty} \frac{1}{2^k} q^{k-1}p$ $=\frac{p}{2}\sum_{k=1}^{\infty} \left(\frac{q}{2}\right)^{k-1}$ $=\frac{p}{2} \frac{1}{1-\frac{q}{2}}$ $=\frac{p}{1+p}$.
-
The PMF of $X$ is given by
\begin{equation}
\nonumber P_X(k) = \left\{
\begin{array}{l l}
pq^{k-1}& \quad \text{for } k=1,2,3,...\\
0 & \quad \text{otherwise}
\end{array} \right.
\end{equation}
where $q=1-p$. Thus,
Problem
If $X \sim Hypergeometric(b,r,k)$, find $EX$.
- Solution
-
The PMF of $X$ is given by
\begin{equation}
\nonumber P_X(x) = \left\{
\begin{array}{l l}
\frac{{b \choose x} {r \choose k-x}}{{b+r \choose k}}& \quad \text{for } x \in R_X\\
0 & \quad \text{otherwise}
\end{array} \right.
\end{equation}
where $R_X=\{\max(0,k-r), \max(0,k-r)+1, \max(0,k-r)+2,..., \min(k,b)\}$. Finding $EX$ directly
seems to be very complicated. So let's try to see if we can find an easier way to find $EX$.
In particular, a powerful tool that we have is linearity of expectation. Can we write $X$
as the sum of simpler random variables $X_i$? To do so, let's remember the random experiment
behind the hypergeometric distribution. You have a bag that contains $b$ blue marbles and $r$
red marbles. You choose $k \leq b+r$ marbles at random (without replacement) and let $X$ be the
number of blue marbles in your sample. In particular, let's define the indicator random variables
$X_i$ as follows:
\begin{equation}
\nonumber X_i = \left\{
\begin{array}{l l}
1 & \quad \text{if the $i$th chosen marble is blue} \\
0 & \quad \text{otherwise}
\end{array} \right.
\end{equation}
Then, we can write
$$X=X_1+X_2+\cdots+X_k.$$
Thus,
$$EX=EX_1+EX_2+\cdots+EX_k.$$
To find $P(X_i=1)$, we note that for any particular $X_i$ all marbles are equally likely to
be chosen. This is because of symmetry: no marble is more likely to be chosen than the $i$th
marble as any other marbles. Therefore,
$$P(X_i=1)=\frac{b}{b+r} \textrm{ for all }i \in \{1,2,\cdots,k\}.$$
We conclude
$EX_i$ $=0 \cdot p(X_i=0)+ 1 \cdot P(X_i=1)$ $=\frac{b}{b+r}$.
Thus, we have $$EX=\frac{kb}{b+r}.$$
-
The PMF of $X$ is given by
\begin{equation}
\nonumber P_X(x) = \left\{
\begin{array}{l l}
\frac{{b \choose x} {r \choose k-x}}{{b+r \choose k}}& \quad \text{for } x \in R_X\\
0 & \quad \text{otherwise}
\end{array} \right.
\end{equation}
where $R_X=\{\max(0,k-r), \max(0,k-r)+1, \max(0,k-r)+2,..., \min(k,b)\}$. Finding $EX$ directly
seems to be very complicated. So let's try to see if we can find an easier way to find $EX$.
In particular, a powerful tool that we have is linearity of expectation. Can we write $X$
as the sum of simpler random variables $X_i$? To do so, let's remember the random experiment
behind the hypergeometric distribution. You have a bag that contains $b$ blue marbles and $r$
red marbles. You choose $k \leq b+r$ marbles at random (without replacement) and let $X$ be the
number of blue marbles in your sample. In particular, let's define the indicator random variables
$X_i$ as follows:
\begin{equation}
\nonumber X_i = \left\{
\begin{array}{l l}
1 & \quad \text{if the $i$th chosen marble is blue} \\
0 & \quad \text{otherwise}
\end{array} \right.
\end{equation}
Then, we can write
$$X=X_1+X_2+\cdots+X_k.$$
Thus,
$$EX=EX_1+EX_2+\cdots+EX_k.$$
To find $P(X_i=1)$, we note that for any particular $X_i$ all marbles are equally likely to
be chosen. This is because of symmetry: no marble is more likely to be chosen than the $i$th
marble as any other marbles. Therefore,
$$P(X_i=1)=\frac{b}{b+r} \textrm{ for all }i \in \{1,2,\cdots,k\}.$$
We conclude
Problem
In Example 3.14 we showed that if $X \sim Binomial(n,p)$, then $EX=np$. We found this by writing $X$ as the sum of $n$ $Bernoulli(p)$ random variables. Now, find $EX$ directly using $EX=\sum_{x_k \in R_X} x_k P_X(x_k)$. Hint: Use $k {n \choose k}=n {n-1 \choose k-1}$.
- Solution
-
First note that we can prove $k {n \choose k}=n {n-1 \choose k-1}$ by the following
combinatorial interpretation: Suppose that from a group of $n$ students we would
like to choose a committee of $k$ students, one of whom is chosen to be the committee
chair. We can do this
- by choosing $k$ people first (in ${n \choose k}$ ways), and then choosing one of them to be the chair ($k$ ways), or
- by choosing the chair first ($n$ possibilities and then choosing $k-1$ students from the remaining $n-1$ students (in ${n-1 \choose k-1}$ ways)).
$EX$ $=\sum_{k=0}^{n} k {n \choose k} p^k q^{n-k}$ $=\sum_{k=1}^{n} k {n \choose k} p^k q^{n-k}$ $=\sum_{k=1}^{n} n {n-1 \choose k-1} p^k q^{n-k}$ $=np\sum_{k=1}^{n} {n-1 \choose k-1} p^{k-1} q^{n-k}$ $=np\sum_{l=0}^{n-1} {n-1 \choose l} p^l q^{(n-1)-l}$ $=np$.
Note that the last line is true because the $\sum_{l=0}^{n-1} {n-1 \choose l} p^l q^{(n-1)-l}$ is equal to $\sum_{l=0}^{n-1} P_Y(l)$ for a random variable $Y$ that has $Binomial(n-1,p)$ distribution, hence it is equal to $1$.
-
First note that we can prove $k {n \choose k}=n {n-1 \choose k-1}$ by the following
combinatorial interpretation: Suppose that from a group of $n$ students we would
like to choose a committee of $k$ students, one of whom is chosen to be the committee
chair. We can do this
Problem
Let $X$ be a discrete random variable with $R_X \subset \{0,1,2,...\}$. Prove $$EX=\sum_{k=0}^{\infty} P(X>k).$$
- Solution
-
Note that
$P(X > 0)$ $=P_X(1)+P_X(2)+P_X(3)+P_X(4)+\cdots$, $P(X > 1)$ $=P_X(2)+P_X(3)+P_X(4)+\cdots$, $P(X > 2)$ $=P_X(3)+P_X(4)+P_X(5)+\cdots$.
Thus$\sum_{k=0}^{\infty} P(X>k)$ $= P(X>0)+P(X>1)+P(X>2)+...$ $=P_X(1)+2P_X(2)+3P_X(3)+4P_X(4)+...$ $=EX$.
-
Note that
Problem
If $X \sim Poisson(\lambda)$, find Var$(X)$.
- Solution
-
We already know $EX=\lambda$, thus Var$(X)=EX^2-\lambda^2$. You can find $EX^2$ directly
using LOTUS; however, it is a little easier to find $E[X(X-1)]$ first. In particular,
using LOTUS we have
$E[X(X-1)]$ $=\sum_{k=0}^{\infty} k(k-1)P_X(k)$ $=\sum_{k=0}^{\infty} k(k-1) e^{-\lambda} \frac{\lambda^k}{k!}$ $=e^{-\lambda} \sum_{k=2}^{\infty} \frac{\lambda^k}{(k-2)!}$ $=e^{-\lambda} \lambda^2 \sum_{k=2}^{\infty} \frac{\lambda^{k-2}}{(k-2)!}$ $=e^{-\lambda} \lambda^2 e^{\lambda}$ $=\lambda^2$.
So, we have $\lambda^2=E[X(X-1)]=EX^2-EX=EX^2-\lambda$. Thus, $EX^2=\lambda^2+\lambda$ and we conclude$\textrm{Var}(X)$ $=EX^2-(EX)^2$ $=\lambda^2+\lambda-\lambda^2$ $=\lambda$.
-
We already know $EX=\lambda$, thus Var$(X)=EX^2-\lambda^2$. You can find $EX^2$ directly
using LOTUS; however, it is a little easier to find $E[X(X-1)]$ first. In particular,
using LOTUS we have
Problem
Let $X$ and $Y$ be two independent random variables. Suppose that we know Var$(2X-Y)=6$ and Var$(X+2Y)=9$. Find Var$(X)$ and Var$(Y)$.
- Solution
- Let's first make sure we understand what Var$(2X-Y)$ and Var$(X+2Y)$ mean. They are Var$(Z)$ and Var$(W)$, where the random variables $Z$ and $W$ are defined as $Z=2X-Y$ and $W=X+2Y$. Since $X$ and $Y$ are independent random variables, then $2X$ and $-Y$ are independent random variables. Also, $X$ and $2Y$ are independent random variables. Thus, by using Equation 3.7, we can write $$\textrm{Var} (2X-Y)=\textrm{Var}(2X)+\textrm{Var(-Y)}=4\textrm{Var}(X)+\textrm{Var(Y)}=6,$$ $$\textrm{Var} (X+2Y)=\textrm{Var}(X)+\textrm{Var(2Y)}=\textrm{Var}(X)+4\textrm{Var(Y)}=9.$$ By solving for Var$(X)$ and Var$(Y)$, we obtain Var$(X)=1$ and Var$(Y)=2$.
