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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

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

set

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

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.

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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

1 Answer 1

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 $npq$.

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