Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What's the meaning of random variables $X_i^2(A)$

For example:

Consider we are doing Bernoulli trials, $\omega =\{A, \text{not} A\}$ with $P(A)=p$ and $P(\text{not} A)=1-p=q$, Given $n$ independent random variables $X_1,X_2,\text{...},x_n$, each taking

$$\begin{align*}X_i(A)=1,X_i(\text{not} A)=0,\end{align*}$$


$$\begin{align*}S_n=\sum _{i=1}^n X_i\end{align*}$$

I can understand this:

$$\begin{align*}E\left(X_k\right)=X_k(A)P(A)+X_k(\text{not} A)P(\text{not} A)=p\end{align*}$$

but feel difficulties in understanding the $X_k^2$ in the variance $V\left(X_k\right)$

$$\begin{align*}V\left(X_k\right)=E\left(X_k^2\right)-\left[E\left(X_k\right)\right]^2\\&=\color{blue}{X_k{}^2(A)P(A)}\color{red}{+}\color{blue}{X_k{}^2(\text{not}A)P(\text{not} A)}-p^2\end{align*}$$

And I can derive the blue part, but how to understand the (real, physical, world, historical...) meaning of $X_k^2(A)$, further, $X_k^3,\text{...}$,

$X_i$ may mean an event, or an event's profit such like earning 10 dollars in one gambling game.

share|cite|improve this question
What's the meaning of $X_i$ to begin with? If that has no meaning, then why should its square have a meaning? – Raskolnikov Jul 22 '13 at 7:10
@Raskolnikov $X_i$ may mean a event, or an event's profit such like earn 10 dollars in one gambling game. – User19912312 Jul 22 '13 at 7:13
I would have $X_i$ here is an indicator random variable – Henry Jul 22 '13 at 7:21
What has important probabilistic meaning is the closely related $(X_i-\mu)^2$, where $\mu$ is the mean of $X_i$. The expectation of $(X_i-\mu)^2$ is the variance of $X_i$, though the (equivalent) formula you quoted is easier to compute. And the variance of $X_i$ is an important measure of the variability of $X_i$. – André Nicolas Jul 22 '13 at 7:23
up vote 3 down vote accepted

$1^2=1$ and $0^2=0$, so if $X_i = 1$ or $0$ then $X_i^2$ is also $1$ or $0$. So in this case $$E[X_i^2] = E[X_i]=\Pr(A)=p.$$

To check your results, the variance of a Bernoulli random variable is $pq$ and of a binomial random variable is $npq$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.