I have two independent multinomial random variables $Y_1$ and $Y_2$. I have to find the distribution of


$$Y_1 - \text{Multinomial}(n_1,(p_1,p_2...p_k))$$ $$Y_2 - \text{Multinomial}(n_2,(p_1,p_2...p_k))$$

I tried using the convolution to calculate the distribution but got stuck after a while

$$P(x_1,x_2..x_k) = \sum_{y_1,y_2..y_n} \binom{n_1}{y_1 y_2..y_k}p_1^{y_1}p_2^{y_2}..p_k^{y_k} \binom{n_2}{(x_1-y_1) (x_2-y_2)..(x_k-y_k)}p_1^{x_1-y_1}p_2^{x_2-y_2}..p_k^{x_k-y_k}$$

such that $y_1+y_2+...+y_n = n_1$ and by similar reasoning we see that $x_1+x_2+...+x_n=n_1+n_2$

$$P(x_1,x_2..x_k) = p_1^{x_1}p_2^{x_2}...p_k^{x_k}\sum_{y_1,y_2..y_n} \binom{n_1}{y_1 y_2..y_k} \binom{n_2}{(x_1-y_1) (x_2-y_2)...(x_k-y_k)}$$

$$P(x_1,x_2..x_k) = (n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}\sum_{y_1,y_2..y_n} \frac{1}{y_1! y_2!..y_k!} \cdot\frac{1}{(x_1-y_1)! (x_2-y_2)!...(x_k-y_k)!}$$

$$P(x_1,x_2..x_k) = \frac{(n_1!)(n_2!) p_1^{x_1}p_2^{x_2}...p_k^{x_k}}{x_1! x_2!..x_k!}\sum_{y_1,y_2..y_n} \binom{x_1}{y_1}\binom{x_2}{y_2}...\binom{x_k}{y_k}$$

But after this I couldn't solve it. Please help

  • $\begingroup$ Are they are independent? $\endgroup$ – Henry Jan 17 at 8:33
  • $\begingroup$ Yeah. They are independent. $\endgroup$ – Sauhard Sharma Jan 17 at 9:04
  • $\begingroup$ Then, as $Y_1$ is the sum of $n_1$ independent $\text{Multinomial}(1,(p_1,p_2...p_k))$ and $Y_2$ is the sum of $n_2$ independent $\text{Multinomial}(1,(p_1,p_2...p_k))$, you find $Y_1+Y_2$ is the sum of $n_1+n_2$ independent $\text{Multinomial}(1,(p_1,p_2...p_k))$ which is $\text{Multinomial}(n_1+n_2,(p_1,p_2...p_k))$ $\endgroup$ – Henry Jan 17 at 10:19
  • $\begingroup$ How can you say that sum of $n_1$ independent Multinomial$(1,(p_1,p_2...p_k))$ is equal to $(n_1,(p_1,p_2...p_k))$. Could you please provide any reference text for this ? $\endgroup$ – Sauhard Sharma Jan 17 at 10:57
  • $\begingroup$ It may depend on your definition of $\text{Multinomial}(n,(p_1,p_2...p_k))$. Wikipedia says "For $n$ independent trials each of which leads to a success for exactly one of $k$ categories, with each category having a given fixed success probability, the multinomial distribution gives the probability of any particular combination of numbers of successes for the various categories" which I would have thought makes my point $\endgroup$ – Henry Jan 17 at 11:03

It would be easier to use characteristic functions. \begin{equation} CF_{\text{Multinomial}(n,(p_1,...,p_k))}(t_1,...,t_k) = \bigg(\sum_{j=1}^k p_je^{it_j}\bigg)^n \end{equation}

As the CF of a sum of random variables is a product of their CFs, it is easy to spot that \begin{equation} X \sim \text{Multinomial}(n_1+n_2,(p_1,p_2...p_k)) \end{equation} as the equality of CFs induces equality of distributions and \begin{equation} CF_X = CF_{Y_1+Y_2} = CF_{Y_1}CF_{Y_2} = \bigg(\sum_{j=1}^k p_je^{it_j}\bigg)^{n_1}\bigg(\sum_{j=1}^k p_je^{it_j}\bigg)^{n_2} = \bigg(\sum_{j=1}^k p_je^{it_j}\bigg)^{n_1 + n_2}= CF_{\text{Multinomial}(n_1 + n_2,(p_1,...,p_k))}(t_1,...,t_k). \end{equation}

  • $\begingroup$ Or we could use moment-generating functions, to avoid complex numbers. Or even better still, we could use probability-generating functions. That has the added benefit of letting us read off the pmf directly afterwards if we'd like to. $\endgroup$ – J.G. Jan 17 at 7:45
  • $\begingroup$ @J.G. Could you please do that and show me ? $\endgroup$ – Sauhard Sharma Jan 17 at 9:09
  • 1
    $\begingroup$ @J.G. What is bad about complex numbers? :-) $\endgroup$ – Math-fun Jan 17 at 9:21
  • $\begingroup$ @SauhardSharma Just replace $e^{it_j}$ with $t_j$. $\endgroup$ – J.G. Jan 17 at 12:16

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.