View Full Version : Binomial distribution confusion.

07-26-2010, 01:31 PM
Hello all.

I am having a bit of confusion taking the conditional expectation of a binomial cumulative distribution. Lets say the probability of a success is x, and the probability of a failure is 1-x. We have n trials.

What is the expected number of successes given that less than k of our n trials ended in a success? Let me give an example for exposition. We flip a fair coin ten times, what is the expected number of heads given that less than 3 heads were flipped?

I suspect the answer is quite straightforward. Any help would be appreciated (I would particularly like a link to a nice website that explains conditional binomial distributions if anybody knows one off hand).

Thanks very much for the help,



07-27-2010, 03:03 AM
One thing you may try is to consider the conditional pmf first.
Actually you can just correct the support and then "normalize" it.

Let X \sim Binomial(n, p)

\forall x \in \{0, 1, ..., n \}
\Pr\{X = x\} = \binom {n} {x} p^{x} (1 - p)^{n - x}

\forall k \in \{1, 2, ..., n+1 \}
\Pr\{X < k\} = \sum_{i=0}^{k-1} \Pr\{X = i\}

\forall x \in \{0, 1, ..., k-1 \}
\Pr\{X = x|X < k\} = \frac {\Pr\{X = x, X < k\}} {\Pr\{X < k\}}
= \frac {\Pr\{X = x\}} {\Pr\{X < k\}}
= \frac {\binom {n} {x} p^{x} (1 - p)^{n - x}}
{\sum_{i=0}^{k-1} \Pr\{X = i\}}

Eventually, we have
E[X|X < k] = \sum_{x=0}^{k-1} x\Pr\{X = x|X < k\}
= \frac {\sum_{x=0}^{k-1} x\Pr\{X = x\}} {\sum_{i=0}^{k-1} \Pr\{X = i\}}

Once you know the original pmf, you can evaluate this with your
calculator easily.