Q:

Finding joint and marginal distributions

Here's a question I got for homework:

A student purchases books for $K$ class hours, where $K$ is a random
  variable with a uniform distribution between $1$ to $3$. The number of
  books the students purchases is also a random variable defined by
  $P(N=n|K=k) = 1/k$, $n = 1,\dots,k$.
What is the joint distribution, what is the marginal probability function of N?

So, first I wrote down $P(K=k): 1/3$, $k=1,2,3$
Now, $P(N=n|K=k) = P(N=n \text{ and } K=k)/P(K=k) = 1/k$, and from that I got $P(N=n\text{ and }K=k) = (1/k)P(K=k)$. When I write down the table of joint distribution the disjoints events didn't sum up to $1$. For example:
\begin{align*}
    P(N=1|K=1) &= 1 \\
    P(N=1|K=2) &= 1/6 \\
    P(N=1|K=3) &= 1/9 \\
\end{align*}
You'll notice that I get P(N=n) = 11/18 for n=1,2,3 And that's what I mean when I say it didn't sum up to 1.
There's obviously something I don't understand, a hint would be great. Where is my mistake?
Thanks!

A:

It's the conditional distributions that don't add to one. Try drawing the joint distribution $P[N=n,K=k]$ in a table for $1 \leq k,n \leq 3$: the nonzero values will form a triangle, where each row and column have sums that make sense. And compare this with the conditional distributions (which are shown first below):
$$
\begin{array}{ccc}
P[N=n|K=k] & n=1 & n=2 & n=3 \\
k=1 & 1 & 0 & 0 \\
k=2 & \frac{1}{2} & \frac{1}{2} & 0 \\
k=3 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3}
\end{array}
$$
$$
\begin{array}{ccc}
P[N=n,K=k] & n=1 & n=2 & n=3 \\
k=1 & \frac{1}{3} & 0 & 0 \\
k=2 & \frac{1}{6} & \frac{1}{6} & 0 \\
k=3 & \frac{1}{9} & \frac{1}{9} & \frac{1}{9}
\end{array}
$$
So you're given $P[K=k]=\frac{1}{3}$ for $1 \leq k \leq 3$ and
$$
P[N=n|K=k]=
\left\{
  \begin{align}
    \frac{1}{k} &\quad \text{for }1 \leq n \leq k;
    \\
    0 &\quad\text{for }n>k.
  \end{align}
\right.
$$
This is the conditional distribution, shown in the first table above.
From this, we can calculate the joint distribution as in the second table:
$P[N=n,K=k]=P[N=n|K=k] \cdot P[K=k] = \frac{1}{3k}$
for $1 \leq n \leq k \leq 3$ (and $0$ for $n>k$).
Finally, $P[N=n]=\sum_{k=n}^{3}P[N=n|K=k]=\sum_{k=n}^{3}\frac{1}{3k}$.
As pointed out by others, the mistake was in writing down the conditional probabilities.

