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So according to the multinomial distribution, the probability function $\Pr(X_1 = x_1, X_2 = x_2, \dots, X_k = x_k)$ is equal to $\dfrac{n!}{x_1! x_2! \cdots x_k!} \cdot p_1^{x_1}\cdot p_2^{x_2} \cdots p_k^{x_k}$

(See http://en.wikipedia.org/wiki/Multinomial_distribution).

And if we were to sum all of the probabilites for all possible values of $x_1, x_2, \dots x_k$, we would get one.

The thing is that summing all of those probablities also includes the situations in which at least one $x_j = 0$. How would I go about computing the following probablity function?:

$$\Pr(X_1=x_1>0, X_2=x_2>0, \ldots, X_k=x_k>0) = ?$$

So basically, the sum of all probabilities of the scenarios in which no $x_j = 0$ and the sum of all the probabilities of the scenarios in which at least one $x_j = 0$ is equal to $1$.

I want to find the sum of all probabilities of the scenarios in which no $x_j = 0$

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  • $\begingroup$ Subtract one from everything and use the multinomial distribution with $n-k$ instead of $n$. $\endgroup$
    – Zach H
    Jul 11, 2013 at 18:25
  • $\begingroup$ Do you mean subtract one from all of the xj's? $\endgroup$
    – Robert Jefferies
    Jul 11, 2013 at 18:26
  • $\begingroup$ You say "The problem is . . .", but that doesn't seem to be a problem at all. If you include all the possible values of $x_1,\ldots,x_n$, all nonnegative integers, satisfying $x_1+\cdots+x_k=n$, including the cases where some of the $x_j$ are $0$, you get $1$. $\endgroup$
    – Michael Hardy
    Jul 11, 2013 at 18:40
  • $\begingroup$ @Zach could you please elaborate a little but more? $\endgroup$
    – Robert Jefferies
    Jul 11, 2013 at 19:04
  • $\begingroup$ @Zach could you please elaborate a little more? $\endgroup$
    – Robert Jefferies
    Jul 12, 2013 at 13:07

1 Answer 1

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You say "The problem is . . .", but that doesn't seem to be a problem at all. If you include all the possible values of $x_1,\ldots,x_n$, all nonnegative integers, satisfying $x_1+\cdots+x_k=n$, including the cases where some of the $x_j$ are $0$, you get $1$.

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  • $\begingroup$ I understand, but I dont want to include the possible values of $x_1, x_2, ... , x_k$ for nonnegative integers, I only want positive integers. The value I should get should be less than 1. $\endgroup$
    – Robert Jefferies
    Jul 11, 2013 at 18:47
  • $\begingroup$ Then why did you say "The problem is . . .", etc.? $\endgroup$
    – Michael Hardy
    Jul 11, 2013 at 18:52
  • $\begingroup$ When I said "The problem is..." I wasn't referring to the actual question at hand, I merely meant that it was a problematic for me... $\endgroup$
    – Robert Jefferies
    Jul 11, 2013 at 18:56
  • $\begingroup$ The real problem is trying to compute the sum of the probabilities for which no $x_j = 0$ $\endgroup$
    – Robert Jefferies
    Jul 11, 2013 at 18:57

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