3 Multivariate Distributions

3.1 Supplementary Concepts

Joint probability mass functions

For discrete random variables $X$ and $Y$, the joint probability mass function, or joint pmf, is

\[\displaystyle p_{X,Y}(x,y)=\Pr[X=x,Y=y]\]

It satisfies

\[\displaystyle p_{X,Y}(x,y)\geq 0\] for every possible pair $(x,y)$, and

\[\displaystyle \sum_x\sum_y p_{X,Y}(x,y)=1\]

The probability that $(X,Y)$ belongs to a set $A$ is

\[\displaystyle \Pr[(X,Y)\in A]=\sum_{(x,y)\in A}p_{X,Y}(x,y).\]

For example,

\[\displaystyle \Pr[X\leq a,Y\leq b]=\sum_{x\leq a}\sum_{y\leq b}p_{X,Y}(x,y).\]

Only pairs belonging to the joint support are included in the summation.

Joint cumulative distribution functions

The joint cumulative distribution function, or joint cdf, of $X$ and $Y$ is

\[\displaystyle {F_{X,Y}(x,y)=\Pr[X\leq x,Y\leq y]}\]

For discrete random variables,

\[\displaystyle F_{X,Y}(x,y) =\sum_{s\leq x}\sum_{t\leq y}p_{X,Y}(s,t).\]

Every joint cdf satisfies

\[\displaystyle 0\leq F_{X,Y}(x,y)\leq 1.\]

It is nondecreasing in each argument and right-continuous in each argument.

It also satisfies

\[\displaystyle \lim_{x\to\infty,;y\to\infty}F_{X,Y}(x,y)=1.\]

The marginal cdfs can be recovered from the joint cdf:

\[\displaystyle F_X(x)=\lim_{y\to\infty}F_{X,Y}(x,y)\]

and

\[\displaystyle F_Y(y)=\lim_{x\to\infty}F_{X,Y}(x,y).\]

Calculating probabilities from a joint distribution

Probabilities involving two discrete random variables are calculated by summing the joint pmf over the appropriate region.

For example,

\[\displaystyle \Pr[a<X\leq b,\;c<Y\leq d]=\sum_{a<x\leq b}\;\sum_{c<y\leq d}p_{X,Y}(x,y).\]

Using the joint cdf,

\[\displaystyle \Pr[a<X\leq b,\;c<Y \leq d]=F_{X,Y}(b,d)-F_{X,Y}(a,d)-F_{X,Y}(b,c)+F_{X,Y}(a,c)\]

This formula extends the univariate cdf difference formula to a rectangular region.

Probabilities over nonrectangular regions are calculated by summing over all pairs satisfying the required condition.

For example,

\[\displaystyle \Pr[X+Y\leq k]=\sum_{x+y\leq k}p_{X,Y}(x,y)\]

Similarly,

\[\displaystyle \Pr[X<Y]=\sum_{x<y}p_{X,Y}(x,y).\]

Marginal probability distributions

A marginal distribution describes one component of a random vector without imposing a condition on the other components.

The marginal pmf of $X$ is obtained by summing over all possible values of $Y$:

\[\displaystyle p_X(x)=\sum_y p_{X,Y}(x,y)\]

Similarly,

\[\displaystyle p_Y(y)=\sum_x p_{X,Y}(x,y)\]

Each marginal pmf satisfies

\[\displaystyle p_X(x)\geq 0\]

and

\[\displaystyle \sum_x p_X(x)=1.\]

Likewise,

\[\displaystyle p_Y(y)\geq 0\]

and

\[\displaystyle \sum_y p_Y(y)=1\]

Marginalization removes one variable from the joint distribution while preserving the probability behavior of the remaining variable.

Conditional probability distributions

The conditional pmf of $X$, given that $Y=y$, is

\[\displaystyle p_{X\mid Y}(x\mid y)=\Pr[X=x\mid Y=y]=\frac{p_{X,Y}(x,y)}{p_Y(y)}\]

provided that

\[\displaystyle p_Y(y)>0\]

Similarly,

\[\displaystyle p_{Y\mid X}(y\mid x)=\frac{p_{X,Y}(x,y)}{p_X(x)}\]

provided that $p_X(x)>0$.

For each fixed value $y$ for which $p_Y(y)>0$,

\[\displaystyle \sum_x p_{X\mid Y}(x\mid y)=1\]

The joint pmf can be reconstructed from a conditional pmf and a marginal pmf:

\[\displaystyle p_{X,Y}(x,y)=p_{X\mid Y}(x\mid y)p_Y(y)\]

It can also be written as

\[\displaystyle p_{X,Y}(x,y)=p_{Y\mid X}(y\mid x)p_X(x)\]

Conditional distributions describe how knowledge of one random variable changes the probability distribution of another.

Independence of random variables

Random variables $X$ and $Y$ are independent if

\[\displaystyle \Pr[X\in A,Y\in B]=\Pr[X\in A] \cdot \Pr[Y\in B]\]

for all relevant sets $A$ and $B$.

For discrete random variables, independence is equivalent to

\[\displaystyle p_{X,Y}(x,y)=p_X(x) \cdot p_Y(y)\]

for every possible pair $(x,y)$.

Independence is also equivalent to

\[\displaystyle p_{X\mid Y}(x\mid y)=p_X(x)\]

whenever $p_Y(y)>0$.

Similarly,

\[\displaystyle p_{Y\mid X}(y\mid x)=p_Y(y)\]

whenever $p_X(x)>0$.

Thus, when $X$ and $Y$ are independent, observing one variable does not change the probability distribution of the other.

If the factorization condition fails for even one pair $(x,y)$, the random variables are not independent.

Expected values of functions of random variables

If $g$ is a function of two discrete random variables, then

\[\displaystyle \operatorname{E}[g(X,Y)]=\sum_x\sum_y g(x,y)p_{X,Y}(x,y)\]

In particular,

\[\displaystyle \operatorname{E}[X]=\sum_x\sum_y x,p_{X,Y}(x,y)\]

and

\[\displaystyle \operatorname{E}[Y]=\sum_x\sum_y y,p_{X,Y}(x,y)\]

These expectations may also be calculated from the marginal pmfs:

\[\displaystyle \operatorname{E}[X]=\sum_x x,p_X(x)\]

and

\[\displaystyle \operatorname{E}[Y]=\sum_y y,p_Y(y)\]

For constants $a$, $b$, and $c$,

\[\displaystyle \operatorname{E}[aX+bY+c]=a\operatorname{E}[X]+b\operatorname{E}[Y]+c\]

Linearity of expectation does not require independence.

Marginal and conditional moments

The $k$th marginal raw moment of $X$ is

\[\displaystyle \operatorname{E}[X^k]=\sum_x x^k p_X(x)\]

The $k$th conditional raw moment of $X$, given $Y=y$, is

\[\displaystyle \operatorname{E}[X^k\mid Y=y]=\sum_x x^k p_{X\mid Y}(x\mid y)\]

In particular, the conditional mean is

\[\displaystyle \operatorname{E}[X\mid Y=y]=\sum_x x,p_{X\mid Y}(x\mid y)\]

The conditional second moment is

\[\displaystyle \operatorname{E}[X^2\mid Y=y]=\sum_x x^2p_{X\mid Y}(x\mid y)\]

The conditional variance is

\[\displaystyle \operatorname{Var}(X\mid Y=y)=\operatorname{E}[X^2\mid Y=y]-\left(\operatorname{E}[X\mid Y=y]\right)^2\]

The conditional standard deviation is

\[\displaystyle \operatorname{SD}(X\mid Y=y)=\sqrt{\operatorname{Var}(X\mid Y=y)}\]

The marginal variance and standard deviation are

\[\displaystyle \operatorname{Var}(X)=\operatorname{E}[X^2]-\left(\operatorname{E}[X]\right)^2\]

and

\[\displaystyle \operatorname{SD}(X)=\sqrt{\operatorname{Var}(X)}\]

Laws of total expectation and total variance

The law of total expectation states that

\[\displaystyle \operatorname{E}[X]=\operatorname{E} \left[\operatorname{E}[X\mid Y]\right]\]

For discrete $Y$,

\[\displaystyle \operatorname{E}[X]=\sum_y\operatorname{E}[X\mid Y=y] \cdot p_Y(y)\]

Thus, the marginal mean is a weighted average of the conditional means.

The law of total variance states that \[\displaystyle \operatorname{Var}(X)=\operatorname{E}[\operatorname{Var}(X\mid Y)]+\operatorname{Var}(\operatorname{E}[X\mid Y])\]

For discrete $Y$,

\[\displaystyle \operatorname{E}[\operatorname{Var}[X\mid Y]]=\sum_y\operatorname{Var}(X\mid Y=y)p_Y(y)\]

The first term measures the average variation within the conditional distributions.

The second term measures the variation among the conditional means.

Together, these terms produce the total marginal variance of $X$.

Covariance and correlation

The covariance of $X$ and $Y$ is

\[\displaystyle \operatorname{Cov}(X,Y)=\operatorname{E}\left[(X-\operatorname{E}[X])(Y-\operatorname{E}[Y])\right]\] The computational form is

\[\displaystyle \operatorname{Cov}(X,Y)=\operatorname{E}[XY]-\operatorname{E}[X]\operatorname{E}[Y]\]

For discrete random variables,

\[\displaystyle \operatorname{E}[XY]=\sum_x\sum_y xy,p_{X,Y}(x,y)\]

A positive covariance indicates that large values of $X$ tend to occur with large values of $Y$.

A negative covariance indicates that large values of $X$ tend to occur with small values of $Y$.

The correlation coefficient is

\[\displaystyle \rho_{X,Y}=\frac{\operatorname{Cov}(X,Y)}{\operatorname{SD}(X)\operatorname{SD}(Y)}\]

provided that both standard deviations are positive.

Correlation satisfies

\[\displaystyle -1\leq\rho_{X,Y}\leq 1.\]

If $X$ and $Y$ are independent and the relevant moments exist, then

\[\displaystyle \operatorname{Cov}(X,Y)=0\]

and

\[\displaystyle \rho_{X,Y}=0.\]

The converse is not generally true. Zero covariance or zero correlation does not necessarily imply independence.

Moments of linear combinations

A linear combination of random variables has the form

\[\displaystyle S=a_1X_1+a_2X_2+\cdots+a_nX_n\]

Its expected value is

\[\displaystyle \operatorname{E}[S]=\sum_{i=1}^n a_i\operatorname{E}[X_i]\]

Its variance is

\[\displaystyle \operatorname{Var}(S)=\sum_{i=1}^n a_i^2\operatorname{Var}(X_i)+2\sum_{i<j}a_i a_j\operatorname{Cov}(X_i,X_j)\]

If the random variables are independent, all covariance terms are zero, so

\[\displaystyle \operatorname{Var}(S)=\sum_{i=1}^n a_i^2\operatorname{Var}(X_i)\]

The standard deviation is

\[\displaystyle \operatorname{SD}(S)=\sqrt{\operatorname{Var}(S)}\]

For the sum

\[\displaystyle S_n=X_1+\cdots+X_n\]

of independent and identically distributed random variables with mean $\mu$ and variance $\sigma^2$,

\[\displaystyle \operatorname{E}[S_n]=n\mu\]

and

\[\displaystyle \operatorname{Var}(S_n)=n\sigma^2\]

For the sample mean

\[\displaystyle \overline{X}=\frac{S_n}{n},\]

\[\displaystyle \operatorname{E}[\overline{X}]=\mu\]

and

\[\displaystyle \operatorname{Var}(\overline{X})=\frac{\sigma^2}{n}.\]

Distributions of sums of independent discrete random variables

Let

\[\displaystyle S=X+Y,\]

where $X$ and $Y$ are independent discrete random variables.

The pmf of $S$ is obtained by convolution:

\[\displaystyle p_S(s)=\sum_x p_X(x)p_Y(s-x)\]

Equivalently,

\[\displaystyle p_S(s)=\sum_{x+y=s}p_X(x)p_Y(y)\]

The probability that the sum belongs to an interval is

\[\displaystyle \Pr[a\leq S\leq b]=\sum_{s=a}^{b}p_S(s),\]

where the sum is taken over possible values of $S$.

For independent Poisson random variables,

\[\displaystyle X\sim\operatorname{Poisson}(\lambda_1)\]

and

\[\displaystyle Y\sim\operatorname{Poisson}(\lambda_2)\]

imply

\[\displaystyle X+Y\sim\operatorname{Poisson}(\lambda_1+\lambda_2)\]

For independent binomial random variables with the same success probability,

\[\displaystyle X\sim\operatorname{Binomial}(n_1,p)\]

and

\[\displaystyle Y\sim\operatorname{Binomial}(n_2,p)\]

imply

\[\displaystyle X+Y\sim\operatorname{Binomial}(n_1+n_2,p)\]

Linear combinations of normal random variables

Suppose $X_1,\ldots,X_n$ are independent normal random variables with

\[\displaystyle X_i\sim N(\mu_i,\sigma_i^2).\]

Then the linear combination

\[\displaystyle S=a_1X_1+\cdots+a_nX_n\]

is also normally distributed:

\[\displaystyle S\sim N\left(\sum_{i=1}^n a_i\mu_i,\sum_{i=1}^n a_i^2\sigma_i^2\right)\]

Thus,

\[\displaystyle \mu_S=\sum_{i=1}^n a_i\mu_i\]

and

\[\displaystyle \sigma_S^2=\sum_{i=1}^n a_i^2\sigma_i^2.\]

Probabilities are calculated by standardization:

\[\displaystyle \Pr[S\leq s]=\Phi\left(\frac{s-\mu_S}{\sigma_S}\right),\]

where $\Phi$ is the standard normal cdf.

More generally, if the normal random variables are jointly normal but not independent, then

\[\displaystyle \operatorname{Var}(S)=\sum_{i=1}^n a_i^2\sigma_i^2+2\sum_{i<j}a_i a_j\operatorname{Cov}(X_i,X_j)\]

A linear combination of jointly normal random variables remains normally distributed.

Order statistics

Let $X_1,\ldots,X_n$ be independent and identically distributed random variables.

The observations arranged from smallest to largest are the order statistics:

\[\displaystyle X_{(1)}\leq X_{(2)}\leq\cdots\leq X_{(n)}.\]

The sample minimum is

\[\displaystyle X_{(1)}=\min(X_1,\ldots,X_n),\]

and the sample maximum is

\[\displaystyle X_{(n)}=\max(X_1,\ldots,X_n).\]

If the common cdf is $F$, then

\[\displaystyle \Pr[X_{(1)}>x]=\Pr[X_1>x,\ldots,X_n>x].\]

By independence,

\[\displaystyle F_{X_{(1)}}(x)=1-\left[1-F(x)\right]^n\]

Similarly,

\[\displaystyle F_{X_{(n)}}(x)=\left[F(x)\right]^n\]

For $a\leq b$,

\[\displaystyle \Pr[X_{(1)}\geq a,;X_{(n)}\leq b]=\Pr[a\leq X_1\leq b,\ldots,a\leq X_n\leq b]\]

Therefore,

\[\displaystyle \Pr[X_{(1)}\geq a,;X_{(n)}\leq b]=\left[\Pr[a\leq X\leq b]\right]^n\]

More generally, for $x\leq y$, the event

\[\displaystyle {X_{(j)}\leq x,;X_{(k)}\leq y}\]

requires at least $j$ observations to be no greater than $x$ and at least $k$ observations to be no greater than $y$.

Its probability can be determined by assigning the observations to the three regions

\[\displaystyle X\leq x,\]

\[\displaystyle x<X\leq y,\]

and

\[\displaystyle X>y,\]

and applying a multinomial probability calculation.

Central Limit Theorem

Let $X_1,\ldots,X_n$ be independent and identically distributed random variables with

\[\displaystyle \operatorname{E}[X_i]=\mu\]

and

\[\displaystyle \operatorname{Var}(X_i)=\sigma^2<\infty.\]

Define the sum

\[\displaystyle S_n=X_1+\cdots+X_n.\]

For sufficiently large $n$, the Central Limit Theorem gives

\[\displaystyle S_n\approx N(n\mu,n\sigma^2)\]

Therefore,

\[\displaystyle \Pr[S_n\leq s]\approx\Phi\left(\frac{s-n\mu}{\sigma\sqrt{n}}\right)\]

For the sample mean

\[\displaystyle \overline{X}=\frac{S_n}{n},\]

the approximation is

\[\displaystyle \overline{X}\approx N\left(\mu,\frac{\sigma^2}{n}\right)\]

Thus,

\[\displaystyle \Pr[\overline{X}\leq x]\approx\Phi\left(\frac{x-\mu}{\sigma/\sqrt{n}}\right).\]

When the underlying random variables are discrete, a continuity correction may improve the normal approximation.

For example,

\[\displaystyle \Pr[S_n\leq k]\]

may be approximated by

\[\displaystyle \Pr[S_n\leq k]\approx\Phi\left(\frac{k+0.5-n\mu}{\sigma\sqrt{n}}\right)\]

Similarly,

\[\displaystyle \Pr[S_n\geq k]\]

may be approximated by

\[\displaystyle \Pr[S_n\geq k]\approx1-\Phi\left(\frac{k-0.5-n\mu}{\sigma\sqrt{n}}\right)\]

The approximation generally improves as $n$ increases and as the underlying distribution becomes less skewed.

3.2 Solved Exercises

Before opening the solutions, take a moment to work through each problem—-your mastery grows most when you engage actively with the mathematics.

Basic probabilities

Let $S(x)$ be defined as follows:

\[s(x) = \left\{ \begin{array}{ll} 1-\dfrac{x}{250} & 0 \leq x < 40 \\ 1-{\left( \dfrac{x}{100} \right)}^2 & 40 \leq x \leq 100 \\ \end{array} \right. \] 1. Calculate the probability that $(25)$ dies within the next $30$ years.
2. Calculate the probability that $(25)$ dies in her fortieth year of live.
3. Calculate the probability that $(25)$ does not die between ages $35$ and $45$
4. Calculate the probability that $(25)$ celebrates her eightieth birthday.

Solution

The probability that $(25)$ dies within the next $30$ years is given by

${_{30}}q_{25}=\dfrac{s(25)-s(55)}{s(25)}=0.225$

The probability $(25)$ dies in her fortieth year of live is given by

${_{14|}}q_{25}={_{14}}p_{25}\;q_{39}=\dfrac{\cancel{s(39)}}{s(25)}\cdot\dfrac{s(39)-s(40)}{\cancel{s(39)}}=0.00\bar{4}$

The probability that $(25)$ does not die between ages $35$ and $45$ is given by

$1-{_{10|10}}q_{25}=1-{_{10}}p_{25}\;{_{10}}q_{35}=1-\dfrac{\cancel{s35}}{s(25)}\cdot\dfrac{s(35)-s(45)}{\cancel{s(35}}=0.930556$

The probability that $(25)$ celebrates her eightieth birthday is given by

$\dfrac{s(80)}{s(25)}=0.4$

$\\$

3.3 Supplementary Exercises

Consider two random variables $X$ and $Y$ with joint probability mass function (PMF) given by

	$Y=0$	$Y=1$	$Y=2$
$X=0$	$\large \frac{1}{6}$	$\large \frac{1}{4}$	$\large \frac{1}{8}$
$X=0$	$\large \frac{1}{8}$	$\large \frac{1}{6}$	$\large \frac{1}{6}$

Find $Pr[X=0, Y \leq 1]$
Find the marginal PMFs of $X$ and $Y$
Find $Pr[Y=1|X=0]$
Are $X$ and $Y$ independent?

Consider the set of points in the grid below.

grid

Find $\mathrm{E}[X\,|\,Y=1\,]$
Find $\mathrm{E}[X\,|\,-1<Y<2\,]$

The number of customers of a Starbucks shop in a given day is $N \sim \mathrm{Poisson}(200)$.

Each customer purchases latte macchiato with probability $0.5$, independently from other customers and independently from the value of $N$. Let $X$ be the number of customers who purchase latte macchiato.

Find $\mathrm{E}[X]$.

The joint probability mass function of the random variables $X$, $Y$, $Z$ is

\[ p(1, 2, 3) = p (2, 1, 1) = p(2, 2, 1)=p(2, 3, 2)=\frac{1}{4} \]

Find $\mathrm{E}[XYZ\,]$
Find $\mathrm{E}[XY + XZ + YZ\,]$

Let $X$ and $Y$ be continuous random variables with the following joint density function (where $c$ is a constant):

\[ f(x,y) = \left\{ \begin{array}{ll} 0.2x+cy, & 0<x<1, 1<y<5 \\ 0, & \mathrm{otherwise} \end{array} \right. \]

Find $\mathrm{Pr}[X+Y>3\,]$

An insurance company is reviewing earthquake damage claims under an industrial property insurance policy.

Let $X$ be the portion of a claim representing damage to the main building, and let $Y$ be the portion of the same claim representing damage to the rest of the property.

The joint density function of $X$ and $Y$ is

\[ f(x,y) = \left\{ \begin{array}{ll} 6[1-(x+y)], & x>0, \,y>0, \,x+y<1 \\ 0, & \mathrm{otherwise} \end{array} \right. \]

Determine the probability that the portion of a claim representing damage to the main building is less than $0.2$.

A fair coin is tossed three times. Let $X$ equal $0$ or $1$ accordingly as a head or a tail occurs on the first toss, and let $Y$ equal the total number of heads that occurs.

Find $\mathrm{Pr}[X=0, Y=2\,]$
Are $X$ and $Y$ independent?
Find $\mathrm{Cov}[X,Y\,]$

A fair coin is tossed three times. Let $X$ equal $0$ or $1$ accordingly as a head or a tail occurs on the first toss, and let $Y$ equal the total number of heads that occurs.

Find $\mathrm{Var}[X+Y\,]$
Is $\mathrm{Var}[X+Y\,]=\mathrm{Var}[X]+\mathrm{Var}[Y]$?

A sample with replacement of size $2$ is randomly selected from the numbers $1$ to $5$.

Let $X=0$ if the first number is even and $X=1$ otherwise; let $Y=1$ if the second number is odd and $Y=0$ otherwise.

Calculate $\mathrm{Pr}[X=0, Y=1\,]$.

A car dealership sells $0$, $1$, or $2$ luxury cars on any day.

When selling a car, the dealer also tries to persuade the customer to buy an extended warranty for the car.

Let $X$ denote the number of luxury cars sold in a given day, and let $Y$ denote the number of extended warranties sold.

Let

$Pr[X=0,Y=0]=\large \frac {1}{6}$
$Pr[X=1,Y=0]=\large \frac {1}{12}$
$Pr[X=1,Y=1]=\large \frac {1}{6}$
$Pr[X=2,Y=0]=\large \frac {1}{12}$
$Pr[X=2,Y=1]=\large \frac {1}{3}$
$Pr[X=2,Y=2]=\large \frac {1}{6}$

Find $Var[X]$.

10 Let $X$ be a random variable with the following distribution, and let $Y=X^2$.

$x$	$-2$	$-1$	$1$	$2$
$\mathbb{P}[X=x]$$	$0.25$	$0.25$	$0.25$	$0.25$

Find $\rho[X,Y\,]$.

The profit for a new product is given by $Z=3X−Y−5$. $X$ and $Y$ are independent random variables with $\mathrm{Var}[X]=1$ and $\mathrm{Var}[Y]=2$.

Find $Var[Z]$.

There are $300$ students in an actuarial faculty, and each student has probability of $0.55$ of passing exam P in the first attempt, independent of any other student.

Let $X$ be the number of students who pass exam P.

Using the normal approximation, what is the probability mass of an interval of two standard deviations of $X$ about the mean?

When a professor asks a question, Charlie gives the correct answer $25\%$ of the time, independent of other questions.

In each class, Charlie may be asked $0$, $1$, or $2$ questions with equal probability.

Let $X$ and $Y$ be the number of questions Charlie is asked, and the number of wrong answers from Charlie in a given class, respectively.

Find $\mathrm{Pr}[X=2, Y=0\,]$.

Consider a transmitter that sends encrypted text messages.

Let $X$ be the travel time (in milliseconds) of a given signal and $Y$ the length of the text (in bytes).

Assume that $Y$ can take only two values, namely $100$ and $10\,000$, and that the following conditional distributions are given

\[ p_{X|Y}(x\,|\,100) = \left\{ \begin{array}{ll} 0.50, & x=0.01 \\ 0.3\bar{3}, & x=0.10 \\ 0.6\bar{6}, & x=1.00 \\ \end{array} \right. \] \[ p_{X|Y}(x\,|\,10\,000) = \left\{ \begin{array}{ll} 0.50, & x=1 \\ 0.3\bar{3}, & x=10 \\ 0.6\bar{6}, & x=100 \\ \end{array} \right. \]

Find $p_{X}(0.01)$.

Let $X_{i}$, $i=1, 2, 3$, be independent Poisson random variables with respective means $\lambda_{i}=i^3$, $i=1, 2, 3$.

Let $X=X_{1}+X_{2}+X_{3}$.

Find $\mathrm{Cov}[X, Y\,]$.

The number of students who enroll in a probability course is a Poisson random variable with mean $100$.

The professor in charge of the course has decided that if the number enrolling is $120$ or more, he will teach the course in two separate sections, whereas if fewer than $120$ enroll, he will teach all of the students together in a single section.

What is the probability that the professor will have to teach two sections?

A fair die is rolled repeatedly.

Let $X$ be the number of rolls needed to obtain a $5$, and $Y$ the number of rolls needed to obtain a $6$.

Calculate $\mathrm{E}[X\,|\,Y=2\,]$.

On a given day, your score in a videogame takes values from the range $101$ to $110$, with probability $0.1$, independent of other days.

Determined to improve your score, you decide to play in three different days, and declare as your score the minimum $X$ of the scores $X_{1}$, $X_{2}$, and $X_{3}$ on the different days.

Calculate $\mathrm{E}[X]$.

In certain city, annual insurance losses due to theft, flooding, and fire are assumed to be independent, exponentially-distributed random variables with respective means $100$, $150$, and $240$.

Determine the probability that the maximum of these losses exceeds $300$.

A diagnostic test for the presence of a disease has two possible outcomes: $1$ for disease present and $0$ for disease not present.

Let $X$ denote the disease state of a patient, and let $Y$ denote the outcome of the diagnostic test.

The joint probability of $X$ and $Y$ is given by:

$\mathrm{Pr}[X=0, Y = 0\,]=0.800$
$\mathrm{Pr}[X=1, Y = 0\,]=0.050$
$\mathrm{Pr}[X=0, Y = 1\,]=0.025$
$\mathrm{Pr}[X=1, Y = 1\,]=0.125$

Calculate $\mathrm{Var}[Y\,|\,X = 1\,]$.

Let $X$ and $Y$ be continuous random variables with joint density

\[ f(x,y) = \left\{ \begin{array}{ll} 24xy, & 0<x<1\,,\,\, 0<y<1-x \\ 0, & x=0.10 \\ \end{array} \right. \]

Calculate $\mathrm{Pr}[Y<X \, | \, X = \frac{1}{3}\,]$

Once a liability claim is reported to an insurance company, the company makes an initial estimate, $X$, of the amount it will pay to the claimant of the liability loss.

When the claim is finally settled, the company pays an amount $Y$ to the claimant.

The company’s actuary has determined that $X$ and $Y$ have the joint density function

\[ f(x,y) = \left\{ \begin{array}{ll} \frac{2}{x^2(x-1)}y^\frac{-(2x-1)}{(x-1)}, & x>1\,,\,\, y>1 \\ 0, & \text{otherwise} \\ \end{array} \right. \]

Given that the initial claim estimated by the company is $2$, determine the probability that the final settlement amount is between $1$ and $3$.

Automobile losses reported to an insurance company are independent and uniformly distributed between $0$ and $20\, 000$.

The company covers each such loss subject to a deductible of $5\, 000$.

Calculate the probability that the total payout on $200$ reported losses is between $1\,000\,000$ and $1\,200\,000$.

A machine consists of two components, whose lifetimes have the joint density function

\[ f(x,y) = \left\{ \begin{array}{ll} 0.02, & x>0\,,\,\, y>0\,,\,\, x+y<10 \\ 0, & \text{otherwise} \\ \end{array} \right. \]

The machine operates until both components fail.

Calculate the expected operational time of the machine.

Annual insurance losses (in millions) due to storm, fire, and theft are assumed to be independent, exponentially distributed random variables with respective means $1.0$, $1.5$, and $2.4$.

Find the probability that the maximum of these losses exceeds $3$.

Two medical plan options will be offered to a company’s employees next year.

An actuary uses the following density function to model the joint distribution of the proportion $X$ of employees who will choose option 1 and the proportion $Y$ who will choose option 2:

\[ f(x,y) = \left\{ \begin{array}{ll} 0.50, & 0<x<0.5\,,\,\, 0<y<0.5 \\ 1.25, & 0<x<0.5\,,\,\, 0<y<1 \\ 1,50, & 0<x<1\,,\,\, 0<y<0.5 \\ 0.75, & 0<x<1\,,\,\, 0<y<1 \\ \end{array} \right. \]

Calculate $\mathrm{Var}[Y\,|\,X=0.75\,]$

Claims filed under auto insurance policies follow a normal distribution with mean $19\,400$ and standard deviation $5\,000$.

What is the probability that the average of 25 randomly selected claims exceeds $20\,000$?

An insurance company issues $1\,250$ insurance policies.

The number of claims filed by a policyholder during one year is a Poisson random variable with mean $2$.

Assume the numbers of claims filed by distinct policyholders are independent of one another.

What is the approximate probability that there is a total of between $2\,450$ and $2\,600$ claims during a one-year period?

A company manufactures electronic components with a lifetime in months that is normally distributed with mean $3$ and variance $1$.

A client buys a number of these components with the intention of replacing them successively as they fail.

The components have independent lifetimes. What is the smallest number of components to be bought so that the succession of components are operational for at least $40$ months with probability at least $0.9772$?

Let $X$ and $Y$ be the number of hours that a randomly selected student watches movies and sporting events, respectively, during a three-month

The following is known about $X$ and $Y$: $\mathrm{E}[X]=50$, $\mathrm{E}[Y]=30$, $\mathrm{Cov}[X, Y]=10$.

One hundred students are randomly selected and observed for these three months.

Let $T$ be the total number of hours that these one hundred students watch movies or sporting events during this three-month period.

Approximate the value of $\mathrm{Pr}[T<7100]$.

	\(Y=0\)	\(Y=1\)	\(Y=2\)
\(X=0\)	\(\large \frac{1}{6}\)	\(\large \frac{1}{4}\)	\(\large \frac{1}{8}\)
\(X=0\)	\(\large \frac{1}{8}\)	\(\large \frac{1}{6}\)	\(\large \frac{1}{6}\)

\(x\)	\(-2\)	\(-1\)	\(1\)	\(2\)
\(\mathbb{P}[X=x]\)$	\(0.25\)	\(0.25\)	\(0.25\)	\(0.25\)