2
$\begingroup$

Let $X$ be a random variable on the probability space $(\Omega,\mathcal B,P)$, with distribution $P_{X}$. Consider the random variable $\hat X$ on the probability space $(\mathbb R,\mathcal B_{\mathbb R},P_{X})$,defined by $\hat X(x)=x$ . Then $P_{\hat X}=P_{X}$.

$\mathcal B $ is $\sigma$-algebra.

thanks for help.

$\endgroup$
3
  • 1
    $\begingroup$ I think you meant the distribution of $\hat{X}$ under $P_X$ is the same as $P_X$. As it is written, $P_{\hat{X}}$ is the distribution of $\hat{X}$ under $P$ which doesn't make sense. $\endgroup$ Nov 14, 2013 at 19:14
  • $\begingroup$ @StefanHansen.Excuse me i take a big mistake and also I'm sorry from peter and delete my comment. $\endgroup$ Nov 15, 2013 at 7:54
  • $\begingroup$ @pualambagher.no problem $\endgroup$
    – peter
    Nov 15, 2013 at 8:01

2 Answers 2

4
$\begingroup$

If $X$ is a random variable on $(\Omega,\mathcal{B},P)$ with distribution $P_X$, and $Y$ defined on $(\mathbb{R},\mathcal{B}(\mathbb{R}),P_X)$ by $Y(x)=x$ for $x\in\mathbb{R}$, then $$ (P_X)_Y=P_X, $$ i.e. the distribution of $Y$ under $P_X$ is exactly $P_X$. This is pretty obvious from the definition of $(P_X)_Y$ which is $$ (P_X)_Y(B)=P_X(Y^{-1}(B)),\quad B\in\mathcal{B}(\mathbb{R}). $$ Now, is it obvious that this equals $P_X(B)$?

$\endgroup$
2
$\begingroup$

$P_X$ and $P_{\hat{X}}$ are probability measures on the Borel sets, so we just need to check that they are equal on arbitrary Borel sets $E$.

So $P_{\hat{X}}(E)$ is defined to be $P_X(\{x\in\mathbb{R}|\hat{X}(x) \in E\})=P_X(\{x\in\mathbb{R}|x \in E\}) = P_X(E)$

$\endgroup$
5
  • $\begingroup$ First and foremost, what "we need to" stress is that $P_{\hat X}$ does not exist. $\endgroup$
    – Did
    Nov 14, 2013 at 19:59
  • $\begingroup$ i'm just taking $P_{\hat{X}}$ to be what it's obviously meant to be. $\endgroup$
    – Louis
    Nov 14, 2013 at 23:55
  • 1
    $\begingroup$ Read better: $\hat X$ is defined on a space endowed with a measure $P_X$, not the measure $P$, hence $(P_X)_{\hat X}$ exists, not $P_{\hat X}$. $\endgroup$
    – Did
    Nov 15, 2013 at 1:47
  • $\begingroup$ you're saying yourself you realize what it's meant to be $\endgroup$
    – Louis
    Nov 15, 2013 at 13:47
  • $\begingroup$ ?? Let us keep it simple: (1) Your post is wrong. (2) To rectify one's post is not something shameful or disgusting. (3) Ergo? $\endgroup$
    – Did
    Nov 15, 2013 at 16:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .