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HW 1 - Solutions
- A fair coin is tossed. If H comes up, then another fair coin is tossed, but if T comes up a fair die is rolled. For each H you win 3 dollars, for each T you lose 3 dollars, and you also win as many dollars as the face of the die shows. Assume that all coin flips and die rolls are independent. This is not a repeated game: there are at most two coin flips and one die roll. Describe a sample space (probability space), together with the probability, on which such an experiment can be modeled. Find the probability mass function of the random variable describing your winnings at the end of this game. Solution: The probability space is S = {HH, HT, T 1 , T 2 , T 3 , T 4 , T 5 , T 6 }
and the probabilities are
P[HH] = P[HT ] =
, P[T 1] = P[T 2] =... P[T 6] =
The random variable we are looking for is given by s = HH HT T 1 T 2 T 3 T 4 T 5 T 6 X(s) = 6 0 − 2 − 1 0 1 2 3 Therefore, the probability mass function is given by i = − 2 − 1 0 1 2 3 6 P(X = i) = 1 / 12 1 / 12 1 / 3 1 / 12 1 / 12 1 / 12 1 / 4
- Choose at random a number N with Poisson distribution with parameter λ > 0. After that, toss a biased coin with probability of H being 1/(N + 1). If H comes up you win 1 if T comes up you lose 1. (a) Compute the probability to get an H or T in this game, and the expected value of your winnings. (b) Compute the probability mass function of N conditioned on getting an H in the coin toss. Solution:
(a) According the the Total Probability Formula,
P[H] =
∑^ ∞
k=
P[H | N = k]P[N = k] =
∑^ ∞
k=
k + 1
e−λ^
λk k!
e−λ λ
∑^ ∞
l=
λl l!
e−λ λ
(eλ−1) =
1 − e−λ λ
Then P[T ] = 1 −
1 − e−λ λ and the expected value is
P[H] − P[T ] =
2 − 2 e−λ^ − λ λ
(b) for k = 0, 1 , 2 ,... , we have
P[N = k | H] =
P[N = k, H] P[H]
P[H | N = k]P[N = k] P[H]
1 k+1 e
−λ λk k! 1 −e−λ λ
eλ^ − 1
λk+ (k + 1)!
Please note that we could have used Bayes’ rule to skip the first equality above.
- Let X 1 ,... , Xm be independent random variables with P[Xi = j] = (^1) n for i = 1,... , m and j = 1,... , n.
(a) Compute P[Xi 6 = Xj , ∀ i, j = 1,... , m, i 6 = j]. (b) Suppose that there are fifty people in a room, and each person’s birthday is uniformly distributed over the 365 days (ignore leap year) of the year and independent of the other peoples birthday. What is the probability that no two people share a birthday. Compute this numerically.
Solution:
(a) Define the events Ei = {Xj 6 = Xk ∀j 6 = k ≤ i} for i = 1,... , k. It then follows from iterated conditioning and the fact that Ej− 1 ⊂ Ej that
P(Em) = P(E 1 )
∏^ m
i=
P(Ei | Ei− 1 )
Now P(E 1 ) = 1 and P(Ei | Ei− 1 ) is the probability that ith number selected is different that than the previous i − 1 numbers given the fact that the previous i − 1 numbers are all different. As there are only n−(i−1) numbers left that Xi can take, so P(Ei | Ei− 1 ) = (n − (i − 1))/n = 1 − (i − 1)/n. Putting this together gives
P(Em) =
m∏− 1
i=
i n
n
n
m − 1 n
Just in case you are not convinced by this argument that P(Ei | Ei− 1 ) = 1 − (i − 1)/n, I will make it a bit more precise. First we need an abstract result: Suppose that (Fn) is a partition of G and P(A | Fn) = p for all p. Then
P(A | G) =
P(A ∩ G)
P(G)
n P(A^ |^ Fn)^ P(Fn) P(G)
p
n P(Fn) P(G)
= p.
Now let Di ⊂ { 1 ,... , n}i^ denote the collection of all sequences of i numbers from the set { 1 ,... , n} which contain no repeats (D for different). Clearly,
Ei =
(d 1 ,d 2 ,...,di)∈Di
{Xj = dj for all 1 ≤ j ≤ i}.
If we fix some vector of different numbers (d 1 , d 2 ,... , di− 1 ) ∈ Di− 1 , then
P
Xi 6 = Xj for all j < i
∣ (^) Xj = dj for all j < i)^ = P(Xi 6 = dj for all j < i) = 1 − (i − 1)/n,
Then
P(Ei | F 1 ∪ F 2 ) = P(Ei) = 1/ 4 , i = 1, 2 , P(E 1 ∩ E 2 | F 1 ∪ F 2 ) = P(E 1 ∩ E 2 ) = 1/ 8 ,
so E 1 and E 2 are not conditionally independent given F 1 ∪ F 2. However,
P(E 1 | F 1 ) = P(E 2 | F 1 ) = 1/ 2 , P(E 1 ∩ E 2 | F 1 ) = 1/ 4 ,
and
P(E 1 | F 2 ) = P(E 2 | F 2 ) = 0, P(E 1 ∩ E 2 | F 2 ) = 0,
so E 1 and E 2 are conditionally independent given only F 1 or only F 2. The intuitive idea here is that E 1 and E 2 share common information about H 2. When we condition, we essentially fix the value of the second coin flip so it is “no longer random.” Once this happens, E 1 only depends upon the first flip and E 2 only depends upon the third flip, so they become (conditionally) independent.
- We have n distinct numbers which are randomly distributed among n players. Assume that all distributions are equally likely. Whenever two players compare their numbers, the player with the highest number is declared the winner. Initially players 1 and 2 compare their numbers; the winner then compares with player 3 and so on. Let X denote the number of times that player 1 is a winner. Compute the probability mass function for X and confirm that it sums to one. Hint: Let Yi denote the number given to the ith player and define the events Ai,m = {Yi ≥ Yj ∀j ≤ m}. Then determine P(Ai,m) for 1 ≤ i ≤ m and P(Ai,m− 1 | Am,m) for 1 ≤ i ≤ m − 1. Solution: First notice that Ai,m is the event that the largest of the first m numbers is given to the ith player. As all distributions of numbers are equally likely, we should have P(Ai,m) = 1/m. Similarly, P(Ai,m− 1 | Am,m) is the likelyhood that the largest of the first m − 1 numbers is given to the ith player, conditional on the fact that the mth players has a larger number than all of them. Well, this conditional information doesn’t make any of the first m − 1 players more or less likely to have the largest number, so we should have P (Ai,m− 1 | Am,m) = 1/(m − 1). If you would prefer a more rigorous argument, let the sample space be the set of all per- mutations of the numbers 1,... , n and assume that all permutations are equally likely. We then notice that exchanging the numbers given to the ith player and jth player produces a bijection which shows that
|Ai,m| = |Aj,m|, i, j = 1,... , m.
and |Ai,m− 1 ∩ Am,m| = |Aj,m− 1 ∩ Am,m|, i, j = 1,... , m − 1 , where |B| denotes the number of points in the set B. As all distributions of the numbers are equally likely, we must have
P(Ai,m) = P(Aj,m), i, j = 1,... , m.
and P(Ai,m− 1 | Am,m) = P(Aj,m− 1 | Am,m), i, j = 1,... , m − 1. But ∪mi=1Ai,m = S where S denotes the sample space, so we may then conclude that
P(Ai,m) = 1/m i = 1,... , m.
and P(Ai,m− 1 | Am,m) = 1/(m − 1), i = 1,... , m − 1.
To compute the distribution of X, we first notice that P(X = 0) = P(A 2 , 2 ) = 1/2 and P(X = n − 1) = P(A 1 ,n) = 1/n. If i = 1,... , n − 2, then {X = i} denotes the event that the first player beats players 2 through i + 1, but loses to player i + 2, so
P(X = i) = P(A 1 ,i+1 ∩ Ai+2,i+2) = P(Ai+2,i+2)P(A 1 ,i+1 | Ai+2,i+2) = 1/(i + 1) × 1 /(i + 2).
We check that this probability mass function sums to one by induction of the number of players. If n = 2, we have P(X = 0) + P(X = 1) = 1/2 + 1/2 = 1. Now assume that the probability mass function sums to 1 for n − 1 players, and write
n∑− 1
i=
P(X = i) = 1/2 +
n∑− 2
i=
(i + 1)(i + 2)
n
n∑− 3
i=
(i + 1)(i + 2)
(n − 1)n
n
n∑− 3
i=
(i + 1)(i + 2)
n − 1
= 1,
where the last equality follows from our inductive assumption that the probability mass func- tion sums to 1 for the case n − 1.
- Suppose we have an urn with 12 balls numbered 1-12 and two dice. Consider the following random variables:
U , the number of the ball that is drawn from the urn, D 1 , the roll of the first die, D 2 , the roll of the second die.
Assume that U , D 1 , D 2 are each uniformly distributed and that these random variables are
