Finding the Expected value and Variance of the Binomial probability distribution

Question

Let $X \sim \text{Binomial}(n,p)$, that is, the probability mass function of $X$, $f(x)$, is such that $$f(x) = \begin{cases} {n \choose x} p^x (1-p)^{n-x} & \text{for } x=0,1,2,\ldots,n\\ 0 & \text{otherwise} \\ \end{cases}$$ I have to find $\mathbb{E}(X)$ and $\mathbb{V}(X)$.

I am fully aware that $\mathbb{E}(x)=np$ but I am supposed to solve this using either a summation or an integral as would be done in any other question. Also, I am aware that $\sum {n \choose k}a^k b^{n-k} = (a+b)^n$ but I'm not sure exactly how to use this since we have a factor of $x$ in our $\mathbb{E}(x)$ function.

So what have you tried? Are you allowed to use a fact of the form that some easy-to-describe random variable has this distribution? — kimchi lover, Nov 08 '17 at 00:56
I edited the question to add something at the end that I was trying to work on — animorphlover3, Nov 08 '17 at 01:00
Can you express $X = X_1 + X_2 + \ldots + X_n$, where the $X_k$ are i.i.d. and each is equal to $1$ with probability $p$ and $0$ with probability $1-p$? — , Nov 08 '17 at 01:11
I'm not sure how that would work, since it is a summation of positive increasing numbers, how can we say that it adds up to 0 or 1. Maybe i'm misunderstanding your question. — animorphlover3, Nov 08 '17 at 01:17
It can be any of the values in ${0, 1, \ldots, n}$, just like your distribution (and with the same probabilities as your distribution). — , Nov 08 '17 at 01:27

score 1 · Accepted Answer · answered Nov 08 '17 at 01:19

1

Note that \begin{align*} \mathbb{E}[X] &= \sum_{x=0}^{n}{x {n \choose x}p^{x}(1-p)^{n-x}} \\ &=\sum_{x=1}^{n}{x {n \choose x}p^{x}(1-p)^{n-x}} & 0 \cdot f(0) = 0 \\ &= \sum_{x=1}^{n}{x \frac{n!}{x!(n-x)!} p^{x}(1-p)^{n-x}} \\ &= np \sum_{x=1}^{n}{\frac{(n-1)!}{(x-1)!(n-1-(x-1))!}p^{x-1}(1-p)^{n-x}} \\ &= np \sum_{x=1}^{n}{{n-1 \choose x-1}p^{x-1}(1-p)^{n-x}} \\ &= np \sum_{y=0}^{n-1}{{n-1 \choose y}p^{y}(1-p)^{(n-1)-y}} & y=x-1 \\ \end{align*} Can you use the binomial theorem here? Can you use the same trick to compute $\mathbb{E}[X(X-1)]$? Note that, since $\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$, you can determine $\mathbb{V}[X]$ from $\mathbb{E}[X(X-1)]$ and $\mathbb{E}[X]$.

answered Nov 08 '17 at 01:19

madprob

2,865

Okay, I see where you're going with this. I get E(x) = (P+1-P)^n = 1^n which doesn't seem right though
Edit: nevermind, I forgot the factor of np
– animorphlover3 Nov 08 '17 at 01:28
Didn't you forget the $np$ outside of the sum? – madprob Nov 08 '17 at 01:30
I'm working on the E(x(x-1)) but i'm not 100% sure how this trick would work for that one. – animorphlover3 Nov 08 '17 at 01:43
You can follow exactly the same steps. However, you have to adapt the reasoning. For example, while $x$ is $0$ only for $x=0$, $x(x-1)$ is $0$ for $x=0$ and $x=1$. – madprob Nov 08 '17 at 01:47
Just a quick clarification, when you pulled your n out of the equation, did you take it out of both the numerator and the denominator? – animorphlover3 Nov 08 '17 at 01:50
I used that $n! = n(n-1)!$ and then pulled $n$ out of the expression. – madprob Nov 08 '17 at 01:52
Okay, I get E(x(x-1))=n^2p^2 which doesn't seem right at all. – animorphlover3 Nov 08 '17 at 01:58
You are almost correct. Probably you used $n! = n^2(n-2)!$, which is incorrect. The correct expression is $n! = n(n-1)(n-2)!$. – madprob Nov 08 '17 at 02:09
That makes more sense, so it's n(n-1)p^2? – animorphlover3 Nov 08 '17 at 02:12
Indeed $E[X(X-1)]= n(n-1)p^2$. – madprob Nov 08 '17 at 02:21
Making Var(x)= n^3p^3-n^2p^3 – animorphlover3 Nov 08 '17 at 02:22
How did you get to this answer? – madprob Nov 08 '17 at 02:26
E(x) * (E(x(x-1)) – animorphlover3 Nov 08 '17 at 02:27
Why do you think $Var[X]=E[X] \cdot E[X(X-1)]$? – madprob Nov 08 '17 at 02:49
looking at it now, I'm seeing that i'm very wrong. I'm just not sure how to find Var(x) from what i've calculated – animorphlover3 Nov 08 '17 at 02:52
Use $Var[X]=E[X^2]-E[X]^2$. Although you haven't computed $E[X^2]$, you can obtain it from other quantities you computed. – madprob Nov 08 '17 at 02:54
I'm thinking it's Var(x) = E(x)+(E(x(x-1)) -(E(x)^2) – animorphlover3 Nov 08 '17 at 03:01
How do you proceed from there? – madprob Nov 08 '17 at 03:04

score 1 · Answer 2 · answered Nov 08 '17 at 01:41

Expanding my comment to an answer, to show an alternative approach.

Let $X_1, X_2, \ldots, X_n$ be i.i.d random variables, each with the distribution $$P(X_k = 1) = p, \qquad P(X_k = 0) = 1-p$$ Define $X = X_1 + X_2 + \ldots + X_n$. Note that for $j \in \{0, 1, \ldots, n\}$, we have $X = j$ if and only if exactly $j$ of the $X_k$'s are equal to $1$. This event has probability $$\begin{aligned} P(X = j)& = P(\text{exactly $j$ of the $X_k$'s equal $1$ and the remaining $n-j$ equal $0$}) \\ &= {n \choose j}p^j (1-p)^{n-j}\\ \end{aligned}$$ which shows that $X$ has the same probability distribution as the one in the OP. We can now easily calculate $E[X]$ and $\operatorname{Var}[X]$ as follows.

First, note that for each $X_k$, we have $$E[X_k] = 1P(X_k = 1) + 0P(X_k = 0) = 1p + 0(1-p) = p$$ and $$E[{X_k}^2] = 1^2P(X_k = 1) + 0^2P(X_k = 0) = 1^2 p + 0^2(1-p) = p$$ and consequently, $$\operatorname{Var}[X_k] = E[{X_j}^2] - E[X_j]^2 = p - p^2$$ Then, since the $X_k$'s are i.i.d., we have $$E[X] = E[X_1] + E[X_2] + \cdots + E[X_n] = nE[X_1] = np$$ and $$\operatorname{Var}[X] = \operatorname{Var}[X_1] + \operatorname{Var}[X_2] + \cdots + \operatorname{Var}[X_n] = n\operatorname{Var}[X_1] = n(p-p^2)$$

Finding the Expected value and Variance of the Binomial probability distribution

2 Answers2