In a sequence of $n$ independent trials the probability of a success at the $i^{\mathrm{th}}$ trial is $p_i$. Find the mean and variance of the total number of successes.

My problem is should I let $X_i$ be the event that the $i^{\mathrm{th}}$ is a success or that $i$ trials have been successful, where $X=X_1+X_2+\cdots +X_n$.

  • $\begingroup$ Your idea is a useful one. We define the random variable $X_i$ by $X_i=1$ if we have a success on the $i$-th trial, and by $X_i=0$ otherwise. Then the number $X$ of successes is given by $X=X_1+\cdots+X_n$. Now $E(X)$ is immediate by the linearity of expectation. For the variance, it will be a good idea to expand $(X_1+\cdots+X_n)^2$. $\endgroup$ Feb 24, 2015 at 17:14
  • $\begingroup$ thank you, I think I can see where to go from here! $\endgroup$
    – guest10923
    Feb 24, 2015 at 17:36
  • $\begingroup$ You are welcome. I thought it best to outline things only, so that you could do the rest. Note that there is a simpler way to get at the variance, since we are dealing with an independent sum. $\endgroup$ Feb 24, 2015 at 17:40
  • $\begingroup$ Hmm, I can't find a way to tidy up the (X1+...+Xn)^2 expression. I tried to use the fact the events are independent therefore E(XY)=E(X)E(Y). I am not sure I know a simpler formula for the variance. $\endgroup$
    – guest10923
    Feb 24, 2015 at 18:00
  • $\begingroup$ The simple way is to use $\text{Var}(X)=\sum \text{Var}(X_i)$. An easy computation (or standard fact) shows that $\text{Var}(X_i)=p_i(1-p_i)$. The harder way is to expand. The mean of $X_i^2$ is $p_i$ since $X_i^2=X_i$. The cross terms have expectation $2\sum_{i\lt j}p_ip_j$. So the expectation of $X^2$ is $\sum p_i+2\sum_{i\lt j}p_ip_j$. Subtract $(E(X))^2$. We get a messy expression that simplifies a lot. $\endgroup$ Feb 24, 2015 at 18:12

1 Answer 1


Recall that for independent random variables $X_i$: $$\mathbb E\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n\mathbb E[X_i] $$ and $$\operatorname{Var}\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n \operatorname{Var}(X_i). $$ Applying these with $\mathbb E[X_i]=p_i$ and $\operatorname{Var}(X_i)=p_i(1-p_i)$ we get that the mean and variance of the sum are $$\sum_{i=1}^n p_i$$ and $$ \sum_{i=1}^n p_i(1-p_i),$$ respectively.


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