If a continuous random variable $X$ has a symmetric distribution around 0, what is the conditional distribution of $X$ given $|X|$?
Tell me more
×
Cross Validated is a question and answer site for
statisticians, data analysts, data miners and data visualization experts. It's 100% free, no registration required.
|
|
|||||||||||||||||||
|
|
Formally, your goal is to find, for each Borel set $A\in\mathscr{R}$, an integrable function $g_A$ (the conditional distribution of $X$ given $|X|$), whose value at $t$ we denote by $$ g_A(t)=:\mathrm{Pr}\left(X\in A\mid |X|=t\right) \, ,$$ satisfying $$ P(X\in A, |X|\in B) = \int_B g_A(t) \, d\mu_{|X|}(t) = \int_B \mathrm{Pr}\left(X\in A\mid |X|=t\right) \, d\mu_{|X|}(t) \, , \qquad (*) $$ for every $B\in\mathscr{R}$, where $\mu_{|X|}$ is the distribution of $|X|$, defined by $\mu_{|X|}(B)=P\left( |X|\in B\right)$, and $$ \mathrm{Pr}\left(X\in {\small\bullet}\mid |X|=t\right) $$ must be a probability measure over $(\mathbb{R},\mathscr{R})$ for almost every $t$ $[\mu_{|X|}]$ (if you can satisfy this for every $t$, then you have a regular version of the conditional distribution). You may guess a candidate $g_A$ thinking as follows. If $t<0$, you don't have to worry: since $\mu_{|X|}$ puts zero mass on negative values, you can define $g_A$ arbitrarily, say, $g_A(t)=1/137.035999$. If $t=0$, you know that $|X|=0$ if and only if $X=0$. So, it is reasonable to try $g_{\{0\}}(0)=1$. If $t>0$, you know that $|X|=t$ if and only if $X=t$ or $X=-t$. Since the distribution of $X$ is symmetric around the origin, it is reasonable to guess that $g_{\{t\}}(t)=g_{\{-t\}}(t)=1/2$. To write your formal proof, you have to show that this candidate $g_A$ satisfies the definition $(*)$. Please, give it a try. |
||||
|
|
