Quizzes

Quiz 12

1. (8 pts) Let $X$ be the number that shows up on the face of a die. Let $Y = X^2$.

   • (6 pts) Compute the correlation coefficient

     $\rho(X,Y) = \frac{\Cov(X,Y)}{\sqrt{\Var(X)} \sqrt{\Var(Y)} }$ Recall that $\Cov(X,Y) = \E[X Y ] - \E[X] \E[Y]$. And recall that

     $$\E[ g(X) ] = \sum_{x} g(x) f_{X}(x)$$

   Solution: Compute

   $$\begin{align} \E[X Y] & = \E[ X^3 ] = \frac16 \sum_{i=1}^6 i^3 = 73.5 \\ \Cov(XY) & = \frac16 \sum_{i=1}^6 i^3 - \left( \frac16 \sum_{i=1}^6 i^2 \right) \left( \frac16\sum_{i=1}^6 i \right) = 73.5 - 15.16 \cdot 3.5 = 20.42 \\ \Var(X) & = \frac16\sum_{i=1}^6 i^2 - \left( \frac16 \sum_{i=1}^6 i \right)^2 = 2.9166 \\ \Var(Y) & = \frac16\sum_{i=1}^6 i^4 - \left( \frac16 \sum_{i=1}^6 i^2 \right)^2 = 149.1389\\ \rho(X,Y) & =\frac{ \frac16 \sum_{i=1}^6 i^3 - \left( \frac16 \sum_{i=1}^6 i^2 \right) \left( \frac16\sum_{i=1}^6 i \right)}{\sqrt{\frac16\sum_{i=1}^6 i^2 - \left( \frac16 \sum_{i=1}^6 i \right)^2} \sqrt{ \frac16\sum_{i=1}^6 i^4 - \left( \frac16 \sum_{i=1}^6 i^2 \right)^2}} = 0.979 \end{align}$$
   • (2 pts) Are $X$ and $Y$ independent? Prove it one way or the other. Hint you can use the correlation coefficient instead of using the definition.

   You couldn’t have calculated the numbers above, but it’s enough to notice that $\rho \neq 0$. This means that $X$ and $Y$ couldn’t have been independent, because if they were $\rho = 0$.

   Alternately, notice $\Prob(X = 1, Y = 1) = \frac16 \neq \Prob(X = 1)\Prob(Y = 1) = \frac16 \cdot \frac16$.

   Some of you thought that $\rho = 0$. If you concluded that $X$ and $Y$ are independent, you get points anyway.

2. (2 pts) Derive a formula for $\Var(X + Y)$ in terms of the variance of $X$, the variance of $Y$ and the covariance of $X$ and $Y$.

   Solution Fully derive

   $$\begin{align} \Var(X + Y) & = \E[ (X+Y)^2 ] - \E[X+Y]^2 \\ & = \E[X^2] + \E[Y^2] - 2 \E[X Y] - \E[X]^2 - \E[Y^2] - 2\E[X] \E[Y] \\ & = \Var(X) + \Var(Y) + 2\Cov(X,Y) \end{align}$$

Quiz 11

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

• (2 pts) Draw a picture of the regions where the pdf $f(x,y)$ is non-zero. What does this tell you about the values that $X$ and $Y$ can take?

  Solution It tells you that $(X,Y)$ can be in the square $[0,1] \times [0,1]$.

  

• (2 pts) Find the marginal densities of $X$ and $Y$.

  Solution This is in your textbook.

  $$f_X(x) = 6x^2 - 4 x^3 \quad f_Y(y) = 2y$$

• (2 pts) Let $A = { X \leq 1/2 }$ and $B = { Y \leq 1/2 }$. Draw pictures of $A$, $B$ and $A \cup B$.

  Solution

  

• (2 pts) Find $\Prob(A), \Prob(B) \AND \Prob( A \cup B)$. What do the rules of probability say about these three quantities you’ve computed?

  Solution We’ll find $\Prob(A), \Prob(B) \AND \Prob(A \cap B)$. There are many other ways of solving this problem.

  $\begin{align} \Prob(A) & = \int_0^{1/2} f_X(x) dx = \left. 2 x^3 - x^2 \right|_0^{1/2} = \frac3{16} \\ \Prob(B) & = \int_0^{1/2} f_Y(y) dx = \left. y^2 \right|_0^{1/2} = \frac1{4} \\ \Prob(A \cap B) & = \int_0^{1/2} \int_{0}^x 4xy dy dx + \int_0^{1/2} \int_{x}^{1/2} 6x^2 dy dx\\ & = \frac{1}{16} \end{align}$ Using the rules of probability, $\Prob(A \cup B) = \Prob(A) + \Prob(B) - \Prob(A \cap B) = \frac3{16} + \frac{4}{16} - \frac1{16} = \frac{6}{16}$

• (2 pts) Find the expected value of $X$. That is, find $\E[X]$.

  $$\E[X] = \int_{0}^{1} x (6x^2 - 4 x^3) dx = \frac{7}{10}$$

Quiz 10

1. (5 pts) Let $X$ be a standard normal random variable and $Z$ a random variable solution of $Z^3 + Z + 1 = X$. Find the probability density function of $Z$.

   Hint: Recall the formula we proved in class. If $X = h(Z)$

   $f_Z(z) = \left|\frac{dh}{dz}(z)\right| f_X(h(z))$ The standard normal density is given by

   $$f_X(x) = e^{-x^2/2} \frac{1}{\sqrt{2\pi}}$$

   Solution The solution may be found in 19.3 (b) in your textbook. You must show (or at least state) that $h = Z^3 + Z + 1$ is one-one to use the theorem. Otherwise you get only two points.

2. (5 pts) We draw two balls with replacement out of an urn in which there are three balls numbered $1,2,3$. Let $X_1$ be the sum of the outcomes and $X_2$ be the product.

   • What’s the joint probability mass function of $f_{X_1,X_2}(x,y)$.

   • What’s the joint distribution function of $(X_1,X_2)$. That is, find

     $$F_{X_1,X_2}(x,y) = \Prob(X_1 \leq x, X_2 \leq y) \quad \forall x,y \in \R$$

   Solution (4pts) for the table. The probability space is $\W = \{ (b_1,b_2) \colon b_i \in \{1,2,3\} \quad i=1,2 \}$ $X_1$ takes values in ${ 2, 3, 4, 5, 6 }$. $X_2$ takes values in ${ 1, 2, 3, 4, 6, 9 }$. Then, we have

   X_1,X_2 1 2 3 4 6 9
   ———- —– —– —– —– —— —– 2 1/9 0 0 0 0 0 3 0 2/9 0 0 0 0
   4 0 0 2/9 1/9 0 0
   5 0 0 0 0 2/9 0
   6 0 0 0 0 0 1/9

   Explain to me (in words or draw a picture) how you’d calculate the cdf, and show me an example or two. If you said, $F(2.0,3.5) = 1/9$ and said “the cdf may be found by summing the probabilities from the mass function table such that $X_1 \leq x$ and $X_2 \leq y$” or some such thing, you’d get points.

Quiz 9

Let $X$ be the number of successes in a sequence of 5000 Bernoulli trials with probability of success $0.4$.

• (3 pts) In terms of the Binomial distribution probabilities, find a representation for $\Prob(X \geq 3000)$ using an infinite sum.

  Solution $\begin{align} \Prob(X \geq 3000) & = \sum_{i=3000}^{5000} \Prob(X = i) \\ & = \sum_{i=3000}^{5000} \binom{5000}{i} 0.4^{i} 0.6^{5000-i} \end{align}$

  There are 2000 terms in the sum that have to be evaluated on the computer. What a mess.

• (3 pts) Estimate $\Prob( X \geq 3000)$ using the central limit theorem. If you use a symbol like $\Phi$ for this, please state the full definition of $\Phi(x)$.

  $$\begin{align} \Prob(X \geq 3000) & = 1 - \Prob(X < 3000)\\ & \approx 1 - \Phi\left( \frac{3000 - 5000 \, 0.4}{\sqrt{ 5000 \, 0.4 \, 0.6}} \right) \end{align}$$

  Another way of writing this that gives the same answer (why?) is as follows.

  $$\begin{align} \Prob(3000 \leq X \leq 5000) & \approx \Phi\left( \frac{5000 - 5000 \, 0.4}{\sqrt{ 5000 \, 0.4 \, 0.6}} \right) - \Phi\left( \frac{3000 - 5000 \, 0.4}{\sqrt{ 5000 \, 0.4 \, 0.6}} \right)\\ \end{align}$$

• (2 pts) What theorem are you using here and why does it apply.

  Central limit theorem. It’s called the Laplace-DeMoivre theorem when it’s specific to the binomial.

Quiz 8

1. Let $f \colon \R \to \R$ be defined by

   $$f(x) = \begin{cases} \frac{c}{1+x^2} & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases}$$

   Does there exist a value of $c$ such that $f$ becomes a probability density function? Justify your answer by showing that there is a $c$ such that

   $$\int_{-\infty}^{\infty} f(x) \, dx = 1$$

   Hints: You might find the following change of variables useful: $x = \tan(\theta)$ and the identity $1+\tan^2(\theta) = \sec^2(\theta) = \cos^{-2}(\theta)$. Also note that $\tan(0) = 0$ and $\tan(\pi/2) = +\infty$.

   Solutions: This is essentially exercise 14.6 in your notes.

Quiz 7

Suppose that 10,000 cars are traveling across a city. One out of 4 cards is gray. Suppose that the probability that any single car has an accident on a particular day is 0.002.

• Let $X$ be the random variable describing the number of gray cars that have an accident. What kind of random variable is it? What are the parameters of the distribution?
• What’s the probability that exactly 3 grey cars have an accident on this day?
• Using the approximation of a Binomial by a Poisson, compute the probability that exactly 3 gray cars have an accident on this day.
• Why can you approximate a Binomial by a Poisson in this case?

Solutions This is exercise 12.2 from the notes.

• $X \sim\Binomial(2,000,0.002)$ since the number of gray cars is $2,000$.
• $\Prob(X = 3) = \binom{2000}{3} 0.002^3 (1-0.002)^{2000-3}$.
• This is in the notes in the appendix with the solutions.
• Using Lemmas 12.3 and Lemma 12.4 in the notes. Basically $n = 2000$ is large.

Quiz 6

1. An urn contains 5 balls, numbered 1 through 5. We draw three of them at random without replacement. Let $X$ be the largest number drawn.

• How big is $\W$? Define $\W$ and $X : \W \to \R$.

• What is the range of values $X$ can take?

• What is the probability mass function of $X$? Hint: Recall that

Solutions:

The solutions may be found in Exercise 11.5 of the notes, with the solutions on page 234.

Note that you are free to choose $\W$ as long as you’re consistent. It might be

The last two questions are answered in the text. Be consistent when computing probabilities whether you choose $\W_1$ or $\W_2$.

Quiz 5

1. Prove the following statements

• If an event $A$ is independent of itself, then $\Prob(A) = 0$ or $\Prob(A) = 1$.
• If $\Prob(A) = 0$ or $\Prob(A) = 1$, then $A$ is independent of any event $B$.

1. Toss a fair coin three times. Let

   $$\begin{align} G_1 & := \{ \text{the 2nd and 3rd tosses give the same outcome} \}.\\ G_2 & := \{ \text{the 1st and 3rd tosses give the same outcome} \}.\\ G_3 & := \{ \text{the 1st and 2nd tosses give the same outcome} \}. \end{align}$$

   Prove that the events are pairwise independent but not independent.

Quiz 4

1. Using the binomial theorem, find the coefficient of $x^6y^{17}$ in

   $$(4x -9y)^{23}$$

   Just give me an unsimplified answer in terms of binomial coefficients, powers of $4$ and $9$, and whatever else may be needed.

   Solution: Using the binomial theorem $\binom{23}{6} 4^6 (-9)^{17}$

2. There are two coins on the table. The first tosses heads with probability $p$ and the second with probability $r$. You select one at random, flip it, and get heads. What is the probability that the second coin was chosen?

   Answer Let $H$ be the event that you get heads. Let $C_2$ be the event that the second coin is chosen. We want $\begin{align} \Prob(C_2 | H) & = \frac{\Prob( H | C_2) \Prob(C_2)}{ \Prob( H | C_1) \Prob(C_1) + \Prob( H | C_2) \Prob(C_2)}\\ & = \frac{r}{ p + r} \end{align}$

Quiz 3

1. A math instructor wanted to divide his classroom of 33 students into groups of 3 for a project. In how many ways can he do this? You must use factorial, permutations or combinations notation. Justify your answer in words.

   Answer: There are two ways to look at this. The first is by combinations. Since there are 33 students, they are to be divided into 11 groups. Label the 11 groups 1 through 11. The number of ways of dividing 33 students into 11 groups of 3 each, where the groups are labeled 1 through 11 is

   $$\binom{33}{3\, 3\, \cdots 3}$$

   But we don’t care if the groups are labeled or not. We needed to label them because that is what the multinomial coefficient measures: it’s the number of ways in which we can divide $n$ into groups of size $3, 3, \ldots 3$, where the groups are labeled. Now is a good time to review the multinomial expansion theorem.

   So the final answer is

   $$\frac{1}{11!} \binom{33}{3\, 3\, \cdots 3}$$

   Another way to look at this is as follows. Out of 33 students, there are

   $$\binom{33}{3}$$

   ways of choosing the first group. Since there are $30$ people left, we may proceed in a similar fashion to determine that we can divide the 33 people into $11$ groups in so many ways:

   $$\binom{33}{3} \binom{30}{3} \cdots \binom{3}{3}$$

   But notice the following. Suppose three people $A,B,C$ are chosen first to form group one. In another configuration, $A,B,C$ might appear in group number $5$. This means that we’re overcounting the number of groups, because the two configurations might be identical, except for the labeling of the groups. To fix this overcounting, we must divide by $11!$, the number of ways in which we can label the $11$ groups. So the final answer is, again,

   $$\frac{1}{11!} \binom{33}{3} \binom{30}{3} \cdots \binom{3}{3}$$

2. Suppose that there are 5 duck hunters, each a perfect shot. A flock of 10 ducks fly over, and each hunter selects one duck at random and shoots. Find the probability that 5 ducks are killed.

   Hint: Your sample space $\W$ consists of all possible duck choices of each of the five hunters. Each hunter chooses one of the $10$ ducks. Your event $A$ is that $5$ different ducks are chosen by the hunters. Compute $

   A

   /

   \W

   $.

   Answer: Let $A$ be the number of ways in which the hunters select $5$ ducks. Clearly,,

   $$|A| = \vphantom{P}^{10} P_5 = \frac{10!}{5!}$$

   This is because the order matters; the hunters are distinct. The total number of ways in which the five hunters can make duck choices is $

   \W

   = 10^5$. Then the probability is

   $$\frac{|A|}{|\W|} = \frac{\vphantom{P}^{10} P_5}{10^5}$$

Quiz 2

1. Prove by induction. Let $A_1,\ldots, A_n$ be $n$ sets in the collection of events $\mathcal{F}$. Then

   $$\Prob( \cup_{i=1}^n A_i ) \leq \sum_{i=1}^n \Prob(A_i)$$

   Hint: Start with the $n=2$ case. That is, for any two steps, show

   $$\Prob(A_1 \cup A_2) \leq \Prob(A_1) + \Prob(A_2)$$

   Solution: The $n=2$ case follows from

   $$\begin{align} \Prob(A_1 \cup A_2) & = \Prob(A_1) + \Prob(A_2) - \Prob(A_1 \cap A_2)\\ & \leq \Prob(A_1) + \Prob(A_2) \end{align}$$

   Once you’ve done this, assume the induction hypothesis: suppose for some number $k$, we can show that for any sets $A_1,\ldots, A_k$,

   $$\Prob( \cup_{i=1}^k A_i ) \leq \sum_{i=1}^k \Prob(A_i)$$

   Your task is to show using the induction hypothesis that for any $k+1$ sets, $A_1,\ldots, A_k$

   $$\Prob( \cup_{i=1}^{k+1} A_i ) \leq \sum_{i=1}^{k+1} \Prob(A_i)$$

   Solution: We have to use the induction hypothesis to show the $n=k+1$ case. Let $A = \cup_{i=1}^k A_i$. Then

   $\begin{align} \Prob( \cup_{i=1}^{k+1} A_i ) & = \Prob(A \cup A_{k+1}) \\ & \leq \Prob(A) + \Prob(A_{k+1}) \\ & \leq \sum_{i=1}^k \Prob(A_i) + \Prob(A_{k+1})\\ & \leq \sum_{i=1}^{k+1} \Prob(A_i) \end{align}$ where we’ve used the $n=2$ case for the second inequality and the $n=k$ induction hypothesis in the third step.

   Let’s see how the induction proof works. We know it’s true for $n=2$ case. So by the induction step with $k=2$, we know that it must be true for $n=3$. We may proceed with the $n=4,\ldots$ and other cases similarly. This proves the statement for any finite integer.

Quiz 1

1. You are a meterologist working for the Philadelphia weather service, and you want to create a probability model the represents the weather over the next two days. On each day, you broadly classify the weather into one of the following three categories: rainy $(R)$ sunny $(S)$ or cloudy $(C)$.

   Describe the space of outcomes $\Omega$ for the weather over the next two days. Please use the letters $R, S$ or $C$ in your answer as shorthand.

   Answer:

   $$\begin{array}{ccc} SS & SR & SC \\ RS & RR & RC \\ CS & CR & CC \end{array}$$

2. Suppose $A$ is the event that it is either rainy or cloudy on at least one of the days. Let $B$ be the event that it is sunny on at least one of the days.

   After running your complicated computer model (codename NTFLX) continuously without any breaks over the long weekend (called a BNGE by the weather industrial complex), you discover that the event $B \setminus A$ is going to occur. What is the event $B \setminus A$ ? Describe it in words.

   Answer:

   $$\begin{align} B & = \{ SS, SC, CS, RS, SR \}\\ A & = \{ RS, RR, RC, CS, CR, CC, SR, SC \}\\ B \setminus A & = \{ SS \} \end{align}$$

   In words, “It’s always sunny in Philadelphia”. At least for the next two days.