Let X and Y be two independent and identically distributed random variables that take only positive integer values. Their PMF is px (n) = py (n) = 2^-n for every n e N, where N is the set of positive integers. 1. Fix at E N. Find the probability P (min{X,Y} <t). Your answer should be a function of t. 2. Find the probability P (X = Y). 3. Find the probability P(X>Y). Hint: Use your answer to the previous part, and symmetry. 4. Fix a positive integer k. Find the probability P (X > KY). Your answer should be a function of k. Hint: You may find the following facts useful. • 2-k = 2-3+1 • For a geometric series, An = aj pn-1, we have m-1 an = a1/(1-r).

1. 1-(1/(2^((2*t)-2)))

2. 1/3
3. 1/3
4. 2/((2^(k+1))-1)

1) 1-(1-CDFx)(1-CDFy) = 1- ((1-CDFx)^2) = 1- ((1-CDFx)^2) = 1-2^(-2t)

integration is not needed as these are discrete RVs

Integration of a^n is a^n/(ln(a)). So why doesn't the answer have ln in it ?

2) Sum[P(Y=y)*P(X=y)] = Sum[1/2^(-2y)] = 1 - 1/2 = 1/2, not sure if this is correct.

why not the 1) (1-(1/2)^t)^2

To find the probability P(min{X,Y} < t), we need to consider the cases where either X or Y is less than t. Since X and Y are independent, we can calculate it as follows:

P(min{X,Y} < t) = 1 - P(min{X,Y} ≥ t)

Now, let's consider the event min{X,Y} ≥ t. This means that both X and Y are greater than or equal to t. Since X and Y are identically distributed, we can calculate this probability for either X or Y and then square it to account for both variables being greater than or equal to t.

P(X ≥ t) = P(Y ≥ t) = ∑(n=t to infinity) P(X = n) = ∑(n=t to infinity) 2^(-n)

Using the formula for the sum of a geometric series, we can simplify this expression:

P(X ≥ t) = P(Y ≥ t) = 2^(-t)/(1 - 2^(-1))

Now, we can calculate P(min{X,Y} < t):

P(min{X,Y} < t) = 1 - [P(X ≥ t)]^2 = 1 - [2^(-t)/(1 - 2^(-1))]^2

Thus, the probability P(min{X,Y} < t) as a function of t is:

P(min{X,Y} < t) = 1 - [2^(-t)/(1 - 2^(-1))]^2

To find the probability P(X = Y), we need to find the probability that X and Y take the same value. This can only happen for positive integer values of n. Thus, for any positive integer n, we have:

P(X = n) = P(Y = n) = 2^(-n)

Since X and Y are independent, we can multiply the probabilities:

P(X = n, Y = n) = P(X = n) * P(Y = n) = (2^(-n))^2 = 2^(-2n)

Now, we can find the probability P(X = Y) by summing over all positive integers n:

P(X = Y) = ∑(n=1 to infinity) P(X = n, Y = n) = ∑(n=1 to infinity) 2^(-2n)

Using the formula for the sum of a geometric series, we can simplify this:

P(X = Y) = ∑(n=1 to infinity) (1/4)^n = (1/4)/(1 - 1/4) = 1/3

Thus, the probability P(X = Y) is 1/3.

To find the probability P(X > Y), we can use the symmetry property of the distribution. Since X and Y are identically distributed, the probability of X being greater than Y is the same as the probability of Y being greater than X. Therefore, we have:

P(X > Y) = P(Y > X)

Now, let's calculate P(Y > X):

P(Y > X) = 1 - P(X ≥ Y)

Using the fact that X and Y are independent, we have:

P(X ≥ Y) = P(X = Y) = 1/3 (as calculated in the previous part)

Therefore,

P(Y > X) = 1 - P(X = Y) = 1 - 1/3 = 2/3

Thus, the probability P(X > Y) is 2/3.

To find the probability P(X > KY), we need to consider the cases where X is greater than KY. Since X is a positive integer and KY is also a positive integer, this means that X must be greater than or equal to K+1. Therefore, we have:

P(X > KY) = P(X ≥ K+1)

Using the geometric series formula, we can calculate P(X ≥ K+1):

P(X ≥ K+1) = ∑(n=K+1 to infinity) P(X = n) = ∑(n=K+1 to infinity) 2^(-n)

Using the formula for the sum of a geometric series, we can simplify this expression:

P(X ≥ K+1) = 2^(-(K+1))/(1 - 2^(-1))

Thus, the probability P(X > KY) as a function of K is:

P(X > KY) = 2^(-(K+1))/(1 - 2^(-1))

Sum[P(Y=y)*P(X=y)] = Sum[2^(-2y)] = 1 - 1/2 = 1/2, not sure if this is correct.

How come I get?

2)1/2
3)1/4