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Two experiments with different probabilities of success run independently $n_1$ and $n_2$ times. I'm modeling the number of successes of each experiment as two independent binomial random variables: $X\sim\mathcal B(n_1, p_1)$ and $Y\sim\mathcal B(n_2, p_2)$.

I would like to know $\Pr[X + Y = k]$ for a constant $k$, i.e., the probability that the sum of the successes of the two experiments in $n_1+n_2$ trials is $k$.

Is there an expression in terms of $n_i$, $p_i$, and $k$ for such probability?

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1 Answer 1

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It will be the convolution: $$\mathsf P(X+Y=k)=\sum_{r=0}^k \mathsf P(X=r)~\mathsf P(Y=k-r)\\=\sum_{r=0}^k \binom{n_1}r\binom{n_2}{k-r}{p_1}^r{p_2}^{k-r}(1-p_1)^{n_1-r}(1-p_2)^{n_2-k+r}$$

Which will not close neatly, unless $p_1=p_2$.

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  • $\begingroup$ thanks for the answer! what do you mean by "close neatly"? $\endgroup$
    – synack
    Oct 21, 2021 at 4:41
  • $\begingroup$ That the series won't have a closed form. However, when $p_1=p_2$ it just ...falls into place neatly.$$\begin{align}\sum_{r=0}^k\binom{n_1}{r}\binom{n_2}{k-r}p^{r}p^{k-r}(1-p)^{n_1-r}(1-p)^{n_2-k+r} &= p^k(1-p)^{n_1+n_2-k}\sum_{r=0}^k\binom{n_1}{r}\binom{n_2}{k-r}\\&=p^k(1-p)^{n_1+n_2-k}\binom{n_1+n_2}{k}\end{align}$$ $\endgroup$
    – Graham Kemp
    Oct 21, 2021 at 7:19
  • $\begingroup$ I see. Thanks. Why isn't the expression in your answer closed? It looks like all the variables are known. $\endgroup$
    – synack
    Oct 21, 2021 at 16:41

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