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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.

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  • $\begingroup$ 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 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 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 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 at 18:30
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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:$$P(X_1+X_2+X_5=8)-P(X_1+X_2=8)-P(X_1+X_5=8)-P(X_2+X_5=8)$$$$+P(X_1=8)+P(X_2=8)+P(X_5=8)=$$$$(p_1+p_2+p_5)^8-(p_1+p_2)^8-(p_1+p_5)^8-(p_2+p_5)^8+p_1^8+p_2^8+p_5^8$$


Addendum:

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.

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