0
$\begingroup$

$X,Y$ are two r.v. $(\Omega,\mathcal{A},\mathbb{P}) \rightarrow (\mathbb{R},\mathcal{B}(\mathbb{R}))$ and have joint density wrt to $\lambda^2$, the two dimensional lebesgue measure.

So $f_X(x) = \int_{\mathbb{R}} f(x,y)\lambda(dy) > 0$ for $\mathbb{P}^X$-a.s. $x \in \mathbb{R}$. And $f_X^{-1}(x)$ ist the density for $\lambda$ absolutely continuous wrt $P^X$.

is then $\int_A \mathbb{P}[X \in dx] = \int_A f_X(x)\lambda(dx)$ ?

Improve this question
$\endgroup$
6
  • $\begingroup$ Fubini: en.wikipedia.org/wiki/… $\endgroup$
    – Zen
    Commented Feb 12, 2015 at 16:37
  • $\begingroup$ $P(X\in A) = P(X\in A, Y\in\mathbb{R}) = \int_{A\times\mathbb{R}} f_{X,Y}(x,y)\,d\lambda^2(x,y) = \int_A\left(\int_\mathbb{R} f_{X,Y}(x,y)\,d\lambda(y)\right)d\lambda(x) = \int_A f_X(x)\,d\lambda(x)$ $\endgroup$
    – Zen
    Commented Feb 12, 2015 at 16:38
  • $\begingroup$ that is not true, this holds: $\int_A \mathbb{P}(X \in dx) \int_B f_{Y|X}(x,y)\lambda(dy) = \int_{A \times B}f(x,y)\lambda^2(dx,dy)$ $\endgroup$
    – user2016445
    Commented Feb 12, 2015 at 17:26
  • $\begingroup$ I wrote four equalities. Which one "doesn't hold"? $\endgroup$
    – Zen
    Commented Feb 12, 2015 at 18:19
  • 1
    $\begingroup$ oh sry my mistake :), $\int_{\mathbb{R}} f_{X,Y}(x,y)d\lambda(y) = 1$ it is true so everythin is fine. thank you it solved my problem. $\endgroup$
    – user2016445
    Commented Feb 12, 2015 at 18:43

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.