0

Let $X$ be a certain random variable with a certain distribution (i.e. $~Bernoulli(X, p)$ where p denotes the probability of $X$). Based on this compute $Variance(X^2)$. Is there a way to imagine what it means to square such a $RV$. For instance, if $X$ describes the outcome of tossing a coin, what would the interpretations for various powers of the $RV$ be?

Thank you in advance!

Samuel

Sam
  • 113
  • 2
    If $X$ takes the values $0,1$ with probability $\frac 1 2 $ each then $X^{n}=X$ for all $n$. – Kavi Rama Murthy Aug 14 '20 at 08:52
  • 1
    Variance is defined by $\mathbf V[X]=\mathbf E[(X-m)^2]=\mathbf E[X^2-2Xm+m^2]=\mathbf E[X^2]-(\mathbf E[X])^2$ where $m=\mathbf E[X]$.

    What could $\mathbf V[X^2]$ mean? Simply $\mathbf V[X^2]=\mathbf E[X^4]-(\mathbf E[X^2])^2=\mathbf E[X^4]-(\mathbf V[X]+m^2)^2$.

    Test with Bernouli $\mathbf V[X^2]=p-(p(1-p)+p^2)^2=p-p^2=p(1-p)=\mathbf V[X]$.

     – Mikael Helin Aug 14 '20 at 09:03
  • 1
    You also, with coin toss, have $\mathbf V[X^n]=\mathbf E[X^{2n}]-(\mathbf E[X^n])^2=p-p^2=\mathbf V[X]$ for any positive $n$. – Mikael Helin Aug 14 '20 at 09:09
  • 1
    If $X$ describes the outcome of tossing a coin then e.g. $X^2$ describes the square of the outcome of tossing a coin. There is nothing more behind it. – drhab Aug 14 '20 at 09:19

0 Answers0