## Expected Value and Variance of a Binomial Distribution

(The Short Way)

Recalling that with regard to the binomial distribution, the probability of seeing $k$ successes in $n$ trials where the probability of success in each trial is $p$ (and $q = 1-p$) is given by
$$P(X=k) = ({}_n C_k) p^k q^{n-k}$$
we can find the expected value and the variance of this probability distribution much more quickly if we appeal to the following properties:

$$E(X + Y) = E(X) + E(Y) \quad \quad \textrm{and} \quad \quad Var(X + Y) = Var(X) + Var(Y)$$
For a random variable $X$ that follows a binomial distribution associated with $n$ trials, probability of success $p$, and probability of failure $q$, let $X_t$ be the random variable that gives the number of successess seen in a single trial (i.e., either $0$ or $1$).

The distribution for $X_t$ is simple in the extreme:
$$\begin{array}{c|c|c}
x & 0 & 1\\\hline
P(X_t = x) & q & p
\end{array}$$

We quickly see that

$$E(X_t) = (0)(q) + (1)(p) = p$$
and
$$\begin{array}{rcl}
Var(X_t) &=& \left[(0^2)(q) + (1^2)(p)\right] - p^2\\
&=& p - p^2\\
&=& p(1-p)\\
&=& pq
\end{array}$$
Now, returning to the expected value of the original random variable $X$ that follows a binomial distribution, note that

$$\begin{array}{rcl}
E(X) &=& \underbrace{E(X_t) + E(X_t) + \cdots E(X_t)}_{\textrm{n terms}}\\
&=& n E(X_t)\\
&=& np
\end{array}$$
Finding the variance of $X$ is just as immediate:

$$\begin{array}{rcl}
Var(X) &=& \underbrace{Var(X_t) + Var(X_t) + \cdots Var(X_t)}_{\textrm{n terms}}\\
&=& n Var(X_t)\\
&=& npq
\end{array}$$
This, of course, immediately gives the standard deviation of $X$:
$$SD(X) = \sqrt{Var(X)} = \sqrt{npq}$$

Somehow, the above justifications seem to fit the simplicity of the results best, don't you think?