Examples

PDF and CDF: Roll 2 dice

Roll two dice colored red and green. Let $X_R$ record the number of dots showing on the red die, $X_G$ the number on the green die, and let $S$ be a random variable giving the total number of dots showing after the roll, namely $S=X_R+X_G$.

• Find the PMFs of $X_R$ and of $X_G$ and of $S$.
• Find the CDF of $S$.
• Find $P[S=8]$.

Solution

(1) Construct sample space:

Denote outcomes with ordered pairs of numbers $(i,j)$, where $i$ is the number showing on the red die and $j$ is the number on the green one.

Therefore $i,j\in \mathbb{N}$ are integers satisfying $1\le i,j\le 6$.

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(2) Create chart of outcomes:

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(3) Define random variables:

We have $X_R(i,j)=i$ and $X_G(i,j)=j$.

Therefore $S(i,j)=i+j$.

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(4) Find PMF of $X_R$:

Use variable $k$ for each possible value of $X_R$, so $k=1,2,\dots ,6$. Find $P_{X_R}(k)$:

$$\begin{array}{c}{P}_{X_R}(k)=P[X_R=k]\\ \\ \gg \gg \frac{|\text{outcomes with }k\text{ on red}|}{|\text{all outcomes}|}\gg \gg \frac{6}{36}=\frac{1}{6}\end{array}$$

Therefore $P_{X_R}(k)=1/6$ for every $k$.

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(5) Find PMF of $X_G$ similarly:

$$P_{X_G}(k)=\frac{1}{6}\text{for all}k$$

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(6) Find PMF of $S$:

$$P_S(k)=P[S=k]\gg \gg \frac{|\text{outcomes with sum }k|}{|\text{all outcomes}|}$$

Count outcomes along diagonal lines in the chart.

$k$	2	3	4	5	6	7	8	9	10	11	12
$P_S(k)$	$\frac{1}{36}$	$\frac{2}{36}$	$\frac{3}{36}$	$\frac{4}{36}$	$\frac{5}{36}$	$\frac{6}{36}$	$\frac{5}{36}$	$\frac{4}{36}$	$\frac{3}{36}$	$\frac{2}{36}$	$\frac{1}{36}$

Evaluate: $P[S=8]=5/36$.

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(7) Find CDF of $S$:

CDF definition: $F_S(x)=P[S\le x]$

Apply definition: add new PMF value at each increment:

$$F_S(x)=\{\begin{array}{cc}0& \phantom{1\le }x<2\\ 1/36& 2\le x<3\\ 3/36& 3\le x<4\\ 6/36& 4\le x<5\\ 10/36& 5\le x<6\\ 15/36& 6\le x<7\\ 21/36& 7\le x<8\\ 26/36& 8\le x<9\\ 30/36& 9\le x<10\\ 33/36& 10\le x<11\\ 35/36& 11\le x<12\\ 36/36& 12\le x\end{array}$$

PMF for total heads count; binomial expansion of 1

A fair coin is flipped $n$ times.

Let $X$ be the random variable that counts the total number of heads in each sequence.

The PMF of $X$ is given by:

$$P_X(k)=\left(\genfrac{}{}{0px}{}{n}{k}\right){\left(\frac{1}{2}\right)}^n$$

Since the total probability must add to 1, we know this formula must hold:

$$\begin{array}{cc}1& =\sum _{\text{possible }k}{P}_X(k)\\ & \\ \gg \gg 1& =\sum _{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right){\left(\frac{1}{2}\right)}^n\end{array}$$

Is this equation really true?

There is another way to view this equation: it is the binomial expansion ${(x+y)}^n$ where $x=\frac{1}{2}$ and $y=\frac{1}{2}$:

$${(\frac{1}{2}+\frac{1}{2})}^n=\sum _{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right){\left(\frac{1}{2}\right)}^n$$

Life insurance payouts

A life insurance company has two clients, $A$ and $B$, each with a policy that pays $100,000 upon death. Consider events $D_1$ that the older client dies next year, and $D_2$ that the younger dies next year. Suppose $P[D_1]=0.10$ and $P[D_2]=0.05$.

Define a random variable $X$ measuring the total money paid out next year in units of $1,000. The possible values for $X$ are 0, 100, 200. Now calculate:

$$\begin{array}{cccc}& P[X=0]\gg \gg & P[D_1^c]P[D_2^c]\gg \gg 0.95\cdot 0.90\gg \gg 0.86& \text{<span class="tml-eqn"></span>}\\ & & & \text{<span class="tml-eqn"></span>}\\ & P[X=100]\gg \gg & 0.05\cdot 0.90+0.95\cdot 0.10\gg \gg 0.14& \text{<span class="tml-eqn"></span>}\\ & & & \text{<span class="tml-eqn"></span>}\\ & P[X=200]\gg \gg & 0.05\cdot 0.10\gg \gg 0.005& \text{<span class="tml-eqn"></span>}\end{array}$$