The problem is as follows:
Alvin shops for probability books for K hours, where K is a random variable that is
equally likely to be 1, 2, 3, or 4. The number of books N that he buys is random and
depends on how long he shops according to the conditional PMF $p_{N|K}(n,k) = \frac{1}{k}$ for n = 1,…,k.
I had found the joint PMF to be $\frac{1}{4k}$ for k = 1,2,3,4 and n = 1,…,k.
However the next question asks to find the marginal PMF of N. Now I would assume that since the joint PMF is independent of N, then if I were to draw a table, each cell in the ith column of k would be the same.
Yet when I check my answer, the marginal PMF is given by $\sum_{k=n}^4 \frac{1}{4k}$, and this does not correspond to my thinking, as it suggests the entry in n=4, k=1 is 0, or is at least not included in the sum.
Can anyone help me understand why this is?
Best Answer
It is perhaps easier to see with the table drawn. The conditional probability table of $P(N|K)$ is given by:
$\:\:\qquad$$k=1$$k=2$$k=3$$k=4$
$n=1$$\hspace{-1pt}\quad1\quad$$\!\quad\frac{1}{2}\!\quad$$\!\quad\frac{1}{3}\!\quad$$\!\quad\frac{1}{4}\!\quad$
$n=2$$\hspace{-1pt}\quad0\quad$$\!\quad\frac{1}{2}\!\quad$$\!\quad\frac{1}{3}\!\quad$$\!\quad\frac{1}{4}\!\quad$
$n=3$$\hspace{-1pt}\quad0\quad$$\hspace{-1pt}\quad0\quad$$\!\quad\frac{1}{3}\!\quad$$\!\quad\frac{1}{4}\!\quad$
$n=4$$\hspace{-1pt}\quad0\quad$$\hspace{-1pt}\quad0\quad$$\hspace{-1pt}\quad0\quad$$\!\quad\frac{1}{4}\!\quad$
and you are somewhat correct in saying that the values of probabilities across the columns are the same, but that is only true for $n\le k$ so beware of the 0 probabilities. I think this is where your confusion arose.
As $K$ has equal probability of taking the values 1, 2, 3 or 4, i.e. $P(K) = \frac{1}{4}$, the table of joint distribution P(N,K) is given by multiplying the table of the conditional probabilities by $\frac{1}{4}$ to give:
$\:\:\qquad$$k=1$$k=2$$k=3$$k=4$
$n=1$$\!\quad\frac{1}{4}\!\quad$$\!\quad\frac{1}{8}\!\quad$$\!\!\quad\frac{1}{12}\!\!\quad$$\!\!\quad\frac{1}{16}\!\!\quad$
$n=2$$\hspace{-1pt}\quad0\quad$$\!\quad\frac{1}{8}\!\quad$$\!\!\quad\frac{1}{12}\!\!\quad$$\!\!\quad\frac{1}{16}\!\!\quad$
$n=3$$\hspace{-1pt}\quad0\quad$$\hspace{-1pt}\quad0\quad$$\!\!\quad\frac{1}{12}\!\!\quad$$\!\!\quad\frac{1}{16}\!\!\quad$
$n=4$$\hspace{-1pt}\quad0\quad$$\hspace{-1pt}\quad0\quad$$\hspace{-1pt}\quad0\quad$$\!\!\quad\frac{1}{16}\!\!\quad$
The marginal probability is simply the sum of the conditional probabilities across the rows, and it is therefore evident that $P(N)=\sum_{k=n}^4 \frac{1}{4k}$.