0
$\begingroup$

I'm wondering how, if it is at all possible, to write the p.d.f. for the following random variable.

Given RVs $X_1$ and $X_2$ distributed according to some joint distribution having known density $p(x_1, x_2)$, if we define the random variable $Y = X_1$ w.p. 1/2 and $Y = X_2$ w.p. 1/2.

what is the p.d.f. for $Y$? is it

$$ p(y) = \frac{1}{2}\left(\int p(y,x) dx + \int p(x,y) dx\right)? $$

More generally, if we have RV $X_1,...,X_n \sim p(\cdot, ..., \cdot)$, with $Y$ chosen uniformly from $X_1,...,X_n$ do we have $$ p(y) = \frac{1}{n}\sum_{i=1}^n \int p(x_1,...,x_{i-1}, y, x_{i+1},...,x_n) dx_1...dx_{i-1}dx_{i+1}...dx_n? $$

$\endgroup$
  • 1
    $\begingroup$ The first formula is correct, the second one misses some integral signs (when added, they will make it correct as well). $\endgroup$ – Did Jul 28 '14 at 22:20
  • $\begingroup$ right! I'm at a loss as to how to prove it correct, the definition of radon-nikodym is rather involved even as a starting point $\endgroup$ – fairidox Jul 28 '14 at 22:22
1
$\begingroup$

In the first case, one can formalize the situation saying that $Y=BX_1+(1-B)X_2$ where $B$ is independent of $(X_1,X_2)$ and $P(B=0)=P(B=1)=\frac12$. Then, for every bounded measurable function $u$, $$E(u(Y))=E(u(Y);B=0)+E(u(Y);B=1)=E(u(X_2);B=0)+E(u(X_1);B=1). $$ By independence, the RHS is $$ E(u(X_2))P(B=0)+E(u(X_1))P(B=1)=\frac12(E(u(X_1))+E(u(X_2)))=\int u(y)g(y)\mathrm dy,$$ where $g=\frac12(f_1+f_2)$, assuming that $X_i$ has density $f_i$. Since this holds for every bounded measurable function $u$, $g$ is the PDF of $Y$. Now, identify each $f_i$ as a marginal of $p$.

$\endgroup$
  • $\begingroup$ ah, yes, thanks! $\endgroup$ – fairidox Jul 28 '14 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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