The expected number of successes in $n$ independent trials with success probability $p$ each is given by the expectation $np$ of the binomial probability $X \sim Binom(n,p)$. Yet you could also write:


Why, intuitively, are these are the same quantity? I'm having a bit of trouble equating these two things and would really appreciate a clean explanation as to why you don't have to think about summing over all possible # of successes that you could get for $n$ trials.

  • $\begingroup$ I don't really understand your question. Do you know what is the expectation by definition? $\endgroup$ – Mesmerized student Oct 6 '15 at 22:54

If you treat the Binomial variable as the sum of $n$ independent, identically distributed Bernoulli trials (i.e. rvs where $X_i = 1$ with probability $p$), then you can use the fact that expectations are linear to state that:

$E[X] = E[\sum{X_i}] = \sum{E[X_i]} = \sum{(1 \times P(X_i=1) + 0 \times P(X_i=0))} = \sum{p} = np$ since you're summing over $n$ events.

That said, you can also prove the expectation directly from that equation you gave (although it's definitely not what I'd call "intuitive").

Start with the fact that the sum of all probabilities of the Binomial rv must be 1, i.e. $\sum_{i=0}^n {{{n}\choose{i}} p^i (1-p)^{n-i}} = 1$.

The derivative of the left side with respect to $p$ is:

$\sum_{i=0}^n {{{n}\choose{i}} (ip^{i-1}(1-p)^{n-i} - (n-i) p^i (n-p)^{n-i-1})} = \frac{1}{p}\sum_{i=0}^n {i{{n}\choose{i}} p^i (1-p)^{n-i}} - \frac{1}{1-p} \sum_{i=0}^n {(n-i){{n}\choose{i}} p^i (1-p)^{n-i}}$

And noting that the derivative of the right side is the derivative of 1, which is equal to 0, you can rearrange the equality of the two to give you what you need, as well.

  • $\begingroup$ Thanks! That does clear it up (but as you mentioned, was definitely less intuitive than I expected). $\endgroup$ – c3200 Oct 7 '15 at 0:02

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