4
$\begingroup$

(Sorry for the title. I has difficulty summarising this problem. I am open to suggestions for a new title.)

Suppose there are a random number of bins of random discrete sizes, each of which contain a random number of balls. Every ball has size 1. The setup is as follows:

  • Bin $i$ has size $S_i$ and contains $K_i$ balls.
  • The total number of balls, $B$, is given by $\sum_{i=1}^N K_i$.
  • The combined volume of all bins, $V$, is given by $\sum_{i=1}^N S_i$.

Assume the following are given:

  • $P(S_i = s)$ for all $s \in \{1, 2, 3,...\}$
  • $P(K_i = k | S_i = s)$ for all $k,s \in \{1, 2, 3,...\}$
  • $P(B = b)$ for all $b \in \{1, 2, 3,...\}$
  • $P(V = v)$ for all $v \in \{1, 2, 3,...\}$

You may also assume that none of the above variables can take the value $0$.

The problem is this. In terms of the above, find an expression for:

  • $P(N=n)$, the probability that the number of bins is $n$.
$\endgroup$
0
$\begingroup$

Not sure if this fit your requirement, but let's try. Consider the pmf of $V$, by law of total probability,

$$ \begin{align} \Pr\{V = v\} &= \sum_{n=1}^v \Pr\{V=v|N=n\}\Pr\{N = n\} \\ &= \sum_{n=1}^v \Pr\left\{\sum_{i=1}^n S_i = v\right\}\Pr\{N = n\} \end{align}$$

for $v = 1, 2, 3, \ldots$ where the second step assumed the indepdendence between $S_i$ and $N$

Note that the LHS is the pmf of $V$ which is given, and in the summand the first term involves the $n$-fold convolution of the given pmf of $S_i$ which can be computed, and only the last term is the desired pmf of $N$ which is unknown yet. So the pmf of $N$ can be solved recursively, say put $v = 1$:

$$ \Pr\{V = 1\} = \Pr\{S_1 = 1\}\Pr\{N = 1\}$$

After solving $\Pr\{N = 1\}$, the remaining can be solved recursively but this maybe computationally heavy.

And the main doubt about this is if the independence assumption is justified in your model. If not, you need to add more dependency assumptions on top of it. You may also consider from the pmf of $B$ if the assumptions are satisfied.

$\endgroup$
  • $\begingroup$ Hmmm I see... yes thank you I'll look over this and see how it works out! $\endgroup$ – JDoe2 Mar 16 '18 at 9:57
  • $\begingroup$ How would you suggest doing this with B instead? $\endgroup$ – JDoe2 Mar 16 '18 at 10:16
  • $\begingroup$ Just replace $V$ by $B$ and $S_i$ by $K_i$ in the above equation, provided that the independence assumption is justified. $\endgroup$ – BGM Mar 16 '18 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.