I have been trying to calculate the probability of getting only a certain subset of events of a multinomial distribution given that a certain number of events occur.

That is, I have a multiniomial distribution

$$ f(x_1, x_2, ... ,x_k; n, p_1, p_2, ... ,p_k) = Pr[X_1 = x_1, X_2 = x_2, ..., X_k = x_k] = \frac{n!}{x_1! x_2!...x_k!} {p_1}^{x_1} {p_2}^{x_2} ... {p_k}^{x_k} $$

I want to calculate, as a example,

$$ Pr[X_1 >0, \mbox{ } X_2 >0, X_5 > 0, \mbox{ } X_3 = 0, \mbox{ } X_4 = 0, \mbox{ } X_6 = 0, \mbox{ } X_7 = 0, ..., X_k = 0 \mbox{ }| n = 8] $$

More generally, I want to be able to calculate the probability that event occurs such that $\{X_i, X_j, X_k\} >0$ and $\{X_m =0 \}$ for $m \not\in \{i,j,k \}$ given $n$.

Is there a way to do this? I only really need to be able to calculate it numerically, but I'm finding even that surprisingly difficult to accomplish.

  • $\begingroup$ I have added something to my answer concerning the general case. It enables you to calculate numerically and is found on base of the inclusion/exclusion principle again. If you want to know more about "how" it was found then please let me know. $\endgroup$
    – drhab
    Jul 10 '19 at 12:28
  • $\begingroup$ Thanks so much. I do a question, if that is alright. Mainly, what does the operator $\wp (S)$ imply in this context? Generally, I would be interested in understanding how to find this expression. $\endgroup$
    – XXXhtd
    Jul 10 '19 at 16:10
  • $\begingroup$ $\wp (S) $ is a notation for the power set of $S $. That is the set of all subsets of $S $. The notation $T\in\wp (S) $ can be replaced by $T\subseteq S $. $\endgroup$
    – drhab
    Jul 10 '19 at 16:51
  • $\begingroup$ Ah, thanks. My set theory is all self taught, so I am very biased in the notation that I'm familiar with. Thanks again! This works perfectly. I'm excited to see if I can figure out the derivation myself!! $\endgroup$
    – XXXhtd
    Jul 10 '19 at 18:30

For a chosen $S\subseteq\{1,2,\dots,k\}$ you want a way to calculate:

$$P(X_i>0\text{ if and only if }i\in S\})$$

or equivalently:$$P(S=\{i\mid X_i>0\})$$

First let me focus on the example you mention where $n=8$ and $S=\{1,2,5\}$.

Then with inclusion/exclusion we find that the probability equals:$$P(X_1+X_2+X_5=8)-P(X_1+X_2=8)-P(X_1+X_5=8)-P(X_2+X_5=8)$$$$+P(X_1=8)+P(X_2=8)+P(X_5=8)=$$$$(p_1+p_2+p_5)^8-(p_1+p_2)^8-(p_1+p_5)^8-(p_2+p_5)^8+p_1^8+p_2^8+p_5^8$$


An expression for the general case is:$$P(\hat S=S)=\sum_{T\in\wp (S)}\left (\sum_{i\in T}p_i\right)^n (-1)^{|S|-|T|}$$where $\hat S$ denotes random set $\{i\mid X_i>0\}$ and $S$ denotes a fixed subset of $\{1,\dots,k\}$.

Finding this starts with the equality:$$P(\hat S=S)=P(\hat S\subseteq S)-P\left(\bigcup_{i\in S}\{\hat S\subseteq S-\{i\}\}\right)$$

where $P(\hat S\subseteq S)=P\left(\sum_{i\in S}X_i=n\right)=\left(\sum_{i\in S}p_i\right)^n$ and for the other term we find an expression by applying the principle of inclusion/exclusion.


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.