Finding the PMF of X conditioned on events

Question

so I know that X= 1/3 y= e^-(5t) (5t)/y=(n)! in order to address the first question PMF would it be P(X|Y)? For the second part, the joint distribution would be the formula fx,y(x|y)(Y)? For the third part, it would be the sum of X?

Answer 1

Your reasoning for the first part is incorrect. To help you understand why, let's try something simpler.

Say I give you a fair six-sided die (numbered from $1$ to $6$). You decide to roll it $n$ times, and count the number of outcomes that are equal to $5$ or $6$. If you were to represent this as a random variable, say $W$, how is $W$ distributed? What are its parameter(s)? What is its probability mass function?

Now turning our attention back to part (a) of the question, note that you are being asked for the distribution of the conditional random variable $X \mid (Y = n)$; that is to say, given that the number of people that this person met in one day is $Y = n$, the random variable $$X \mid (Y = n)$$ counts the number of those people that were infected. This is precisely the same situation as the die roll: given that you chose the roll the die $n$ times, and the probability that you get a $5$ or $6$ in any single roll is $p = 2/6 = 1/3$, the random number of such rolls is analogous to the number of infected people among the $n$ people that the person encountered that day. Another way to see this is that $X | Y$ cannot be arbitrarily large. Once you set in advance the number of encounters, the total number infected cannot exceed the total number of encounters.

For part (b), you simply use the fact that $$\Pr[(X = x) \cap (Y = y)] = \Pr[X = x \mid Y = y]\Pr[Y = y].$$ This arises from the definition of conditional probability.

For part (c), one uses the law of total probability: $$\Pr[X = x] = \sum_{y=0}^\infty \Pr[(X = x) \cap (Y = y)].$$ The goal of this part of the exercise is to show that the unconditional distribution of the total number of people infected in a given day is itself Poisson. This is in contrast to the conditional distribution you computed in part (a), because in this case, you are not told in advance how many encounters occurred. The basic intuition behind this is as follows: to find the unconditional probability, you take a weighted sum of all of the conditional probabilities, where the weights are the probabilities of observing each possible number of encounters.

To give you a simpler example, let's return to the die roll. Originally, you computed a PMF for $W$ that assumes $n$ is fixed in advance. Now suppose instead I give you a fair coin in addition to the die, and I tell you to flip the coin before you roll the die. If you get heads, you roll the die $4$ times. If you get tails, you roll the die $2$ times. Now, before you have flipped the coin, what is the probability distribution for $W$?