Question 1
People are arriving at a party one at a time. While waiting for more people to arrive
they entertain themselves by comparing their birthdays. Let X be the number of people
needed to obtain a birthday match, i.e., before person X arrives there are no two people
with the same birthday, but when person X arrives there is a match. Find the PMF of
X.
Question
$$
\dfrac{x-1}{365} \prod_{i=1}^{363} \dfrac{365-i}{365}
$$
Where \(x\) is between 2 and 365
There are simpler ways to write this without sigma notation, I will leave that to the readers to figure out (not because I am lazy)
Answer
P(person number x is the first person to have same birthday) = P(noone has same birthday)P(last person has same birthday as x-1 pervious)
Explanation
Question 2
(a) Independent Bernoulli trials are performed, with probability 1/2 of success, until
there has been at least one success. Find the PMF of the number of trials performed.
(b) Independent Bernoulli trials are performed, with probability 1/2 of success, until
there has been at least one success and at least one failure. Find the PMF of the number
of trials performed.
Question
A)
$$
p_X(x) = P(X = x) = \dfrac{1}{2^x}
$$
Where \(x\) is the number of bernouli trials before a success
B)
$$
p_X(x) = P(X = x) = \begin{cases}
\dfrac{x-1}{2^{(x-1)}} & \text{if} x > 1 \\
0 & \text{otherwise}
\end{cases}
$$
Answer
A) Count the number of ways you can flip a coin \(x\) times and have the first \(x-1\) flips be tails and the last flip be heads. There is only one way to do this but there are \(2^x\) possible other ways for the coin to land.
B) Count the number of ways that you can have either all heads and the last flip is tails or vis-versa with at least two flips.
Explanation
Question 3
Let X be an r.v. with CDF F , and Y = μ + σX, where μ and σ are real numbers with
σ > 0. (Then Y is called a location-scale transformation of X; we will encounter this
concept many times in Chapter 5 and beyond.) Find the CDF of Y , in terms of F
Question
$$
F_Y(y) = F_X(\dfrac{y-\mu}{\sigma})
$$
Answer
When it asks us to find Y in terms of F, it is asking us to define a new function \(F_Y\) that uses our new variable \(Y\) instead of \(X\).
First write \(F\).
$$
F_X(x) = P(X = x)
$$
now write out \(F_Y\)
$$
F_Y(y) = P(Y = y)
$$
Now sub \(Y\) with \(Y = \mu + \sigma X\)
$$
F_Y(y) = P(\mu + \sigma X = y)
$$
And re-write with \(F\)
$$
F_Y(y) = F_X(\dfrac{y- \mu}{\sigma})
$$
Explanation
Question 4
Let n be a positive integer and
$$
F(x) = \dfrac{\lfloor x \rfloor}{n}
$$
for \(0 \le x \le n, F (x) = 0\) for x \(\lt\) 0, and F (x) = 1 for x \(\gt\) n, where \(\lfloor x \rfloor\) is the greatest
integer less than or equal to x. Show that F is a CDF, and find the PMF that it
corresponds to.
Question
We can prove that \(\dfrac{\lfloor x \rfloor }{n}\) is a CDF because it is non-decreasing and it sums to 1.
We can find its PDF by finding each value for \(F(x) - F(x-1)\). Using the above function,
$$
F(0) = 0
$$
$$ F(1) - F(0) = \frac{1}{n}
$$
$$
F(2) - F(1) = \frac{1}{n}
$$
therefore, the PMF of \(F(x) = \dfrac{\lfloor x \rfloor}{n}\) is \(\dfrAC{1}{n}\)
Answer
Question 5
(a) Show that p(n) = \((\frac{1}{2})^{n+1})\)
for n = 0, 1, 2, . . . is a valid PMF for a discrete r.v.
(b)
Find the CDF of a random variable with the PMF from (a).
Question
A) We can show that this is a PMF by seeing that the limit when n = \(\infty\) is 1 and that all values in the given domain are non-negative
$$
\lim_{n \to \infty} \left( \frac{1}{2} \right)^{n+1} = 1
$$
B) We get the CDF by adding all values from \(0\) to \(n\)
$$
\sum_{i=0}^{n} \dfrac{1}{2^{n+1}} = \dfrac{2^{n+1}-1}{2^{n+1}}
$$
Answer
Question 7
Bob is playing a video game that has 7 levels. He starts at level 1, and has probability
p1 of reaching level 2. In general, given that he reaches level j, he has probability \(p_j\) of
reaching level \(j + 1\), for \(1 \le j \le 6\). Let X be the highest level that he reaches. Find the
PMF of X (in terms of \(p_1\), \(\ldots\) , \(p_6\)).
Question
$$
p_X(x) = \begin{cases}
1-p_1 & \text{if } x = 1 \\
(1-p_{x+1}) \prod_{i=1}^{x} p_j & \text{if } x \in \{2,3,4,5,6\} \\
\prod_{i=1}^{x} p_j & \text{if } x = 7 \\
0 & \text{otherwise}
\end{cases}
$$
Answer
The probability that the highest level the highest level he reaced is \(j\), that means he reached level \(j\) but not level \(j+1\). There are several cases because he could reach the last level or only the first (lose right away). We also want to include the implicit case that X is not in the domain.
Explanation
Question 8
There are 100 prizes, with one worth $1, one worth $2, \(\ldots\) , and one worth $100. There
are 100 boxes, each of which contains one of the prizes. You get 5 prizes by picking random boxes one at a time, without replacement. Find the PMF of how much your
most valuable prize is worth (as a simple expression in terms of binomial coefficients).
Question
$$
P_X(X = x) = f(x) = \dfrac{ {x \choose 5}5! }{100 \choose 5}
$$
Answer
There are \(5\) numbers being selected, therefore we have \(5!\)
There are \({x \choose 5}\) ways for \(x\) to be the highest number selected.
There are \({100 \choose 5}\) total ways to select \(5\) numbers of \(100\).
Explanation