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

$$X=Y_1+Y_2$$

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

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

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  • $\begingroup$ Are they are independent? $\endgroup$ – Henry Jan 17 '19 at 8:33
  • $\begingroup$ Yeah. They are independent. $\endgroup$ – Sauhard Sharma Jan 17 '19 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 '19 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 '19 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 '19 at 11:03
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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}

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  • $\begingroup$ 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 '19 at 7:45
  • $\begingroup$ @J.G. Could you please do that and show me ? $\endgroup$ – Sauhard Sharma Jan 17 '19 at 9:09
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    $\begingroup$ @J.G. What is bad about complex numbers? :-) $\endgroup$ – Math-fun Jan 17 '19 at 9:21
  • $\begingroup$ @SauhardSharma Just replace $e^{it_j}$ with $t_j$. $\endgroup$ – J.G. Jan 17 '19 at 12:16

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