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
Best Answer
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}