Question

Consider \(n\) multinomial trials, where each trial independently results in outcome \(i\) with probability \(p_{i}, \sum_{i=1}^{k} p_{i}=1 .\) With \(X_{i}\) equal to the number of trials that result in outcome \(i\), find \(E\left[X_{1} \mid X_{2}>0\right]\)

Short answer

The expected number of trials resulting in outcome 1 given that there's at least one trial resulting in outcome 2 is given by: $$E[X_1\mid X_2 >0] = \frac{\sum_{x_1=0}^{n-1} \sum_{x_2=1}^{n-x_1} x_1 \frac{n!}{x_1! x_2! (n-x_1-x_2)!} p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}}{1-p_1^n}$$

Step-by-step solution

In a multinomial distribution, the joint PMF of (\(X_1\), \(X_2\)) is given by: $$P(X_1=x_1, X_2=x_2) = \frac{n!}{x_1! x_2! (n-x_1-x_2)!} p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}$$ Here, we consider only the cases where \(0\leq x_1 \leq n - 1\) and \(0\leq x_2 \leq n - x_1\), since \(X_1 + X_2 \leq n\). The rest of the trials will fall into outcomes 3,4,5,...,k. #Step 2: Find \(E[X_1 \mathbb{1}_{\{X_2 > 0\}}]\)#

In order to find \(E[X_1\mathbb{1}_{\{X_2>0\}}]\), we'll first compute \(x_1 P(X_1=x_1, X_2=x_2)\) for each possible value of \(x_1\) and \(x_2\) where \(x_2 > 0\), and then sum over all possibilities: $$E[X_1\mathbb{1}_{\{X_2>0\}}]= \sum_{x_1=0}^{n-1} \sum_{x_2=1}^{n-x_1} x_1 P(X_1=x_1, X_2=x_2) = \sum_{x_1=0}^{n-1} \sum_{x_2=1}^{n-x_1} x_1 \frac{n!}{x_1! x_2! (n-x_1-x_2)!} p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}$$ #Step 3: Find \(P(X_2 > 0)\)#

Next, we'll compute the probability \(P(X_2 > 0)\) which means at least one trial will result in outcome 2. We can calculate this as the complement: \(P(X_2 > 0) = 1 - P(X_2 = 0)\). Since all outcomes are independent, we have: $$P(X_2=0) = n!\frac{p_1^{n-x_2} p_3^{x_2}}{(n-x_2)!x_2!}$$ Substitute \(x_2=0\) to the above equation and subtract from 1: $$P(X_2 > 0) = 1 - n!\frac{p_1^n}{n!} = 1 - p_1^n$$ #Step 4: Calculate \(E[X_1 | X_2 > 0]\)#

Finally, we can use the definition of conditional expectation to find the desired expectation: $$E[X_1\mid X_2 >0] = \frac{E[X_1\mathbb{1}_{\{X_2>0\}}]}{P(X_2>0)}= \frac{\sum_{x_1=0}^{n-1} \sum_{x_2=1}^{n-x_1} x_1 \frac{n!}{x_1! x_2! (n-x_1-x_2)!} p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}}{1-p_1^n}$$ This expression gives us the expected number of trials resulting in outcome 1 given that there's at least one trial resulting in outcome 2.