2
$\begingroup$

I'm completely stuck on a problem and maybe someone of you out there has some thoughts on how I could proceed.

The setting is the following: Let $f: \mathbb{R}^2 \to \mathbb{R},(x,y) \mapsto f(x,y)$ be a continuous function. Next consider two iid random variables $X,Y$ on a probability space $(\Omega,\mathcal{A}, \mathbb{P}$). Then we could "randomize" $f$ via $$f(X,Y): \Omega \to \mathbb{R}, \omega \to f(X(\omega),Y(\omega)).$$

(We always identify $\mathbb{R}$ with the Borel-sigma-algebra as measurable space). Next let $a \in Y(\Omega)$ be arbitrary and set $$f(X,a): \Omega \to \mathbb{R}, \omega \to f(X(\omega),a).$$

Then, since $X$ and $Y$ are independent we can obtain the distributional equality $$\mathbb{P}^{f(X,a)} = \mathbb{P}^{f(X,Y) \mid Y = a}.$$

I hope, that everything sounds ok so far.

Now, further let $L(a) = \mathbb{E}[(f(X,a))^2] = \Vert f(X,a) \Vert_{L^2}^2$. In other words $L(\cdot): Y(\Omega) \to \mathbb{R}, a \mapsto L(a)$. Hence we could also "randomize" this function as $$L(Y): \Omega \to \mathbb{R}, \omega \mapsto L(Y(\omega)).$$

Now, I tried to compute the expectation of $L(Y)$ in terms of the expressions, but here is where I'm stuck at the moment. My idea was the following: $$\mathbb{E}(L(Y)) = \int_\Omega L(Y(\omega)) d\mathbb{P}(\omega) \\ = \int_\Omega \mathbb{E}[(f(X,Y(\omega))^2] d\mathbb{P}(\omega) \\ = \int_\Omega \int_\Omega (f(X(\tau), Y(\omega))^2 d\mathbb{P}(\tau)d\mathbb{P}(\omega).$$

From here on my plan was to proceed with Fubini's Theorem to change the order of integration, but to me it doesn't seem right, that $\tau$ and $\omega$ could be non-equal (since I defined $f(X,Y)$ such that $X,Y$ always have the same argument). Did I mess something up?

Or am I on the right path and could define the random variable $$f'(X,Y) : \Omega\times\Omega \to \mathbb{R}, (\tau,\omega) \mapsto f(X(\tau), Y(\omega))$$ on $(\Omega\times\Omega, \mathcal{A} \otimes \mathcal{A}, \mathbb{P} \times \mathbb{P})$ and then $\mathbb{E}(L(Y)) = \mathbb{E}(f'(X,Y))$? But then I'm still stuck, since I have no clue how I could connect $f'(X,Y)$ and $f(X,Y)$.

Maybe I'm just being a bit clumsy and things are actually much simpler. In any case, I would be grateful for any help.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

The law of total expectation (a.k.a. law of iterated expectation, tower property, etc.) tells that

$$\mathbf{E}[L(Y)]=\mathbf{E}[\mathbf{E}[f(X,Y)^2\mid Y]]=\mathbf{E}[f(X,Y)^2].$$

Alternatively, sticking to the integral notation, we have

\begin{align*} \mathbf{E}[L(Y)] &= \iint_{\Omega^2} f(X(\tau),Y(\omega))^2 \, \mathbf{P}(\mathrm{d}\tau)\mathbf{P}(\mathrm{d}\omega) \\ &= \iint_{\mathbb{R}^2} f(x, y)^2 \, \mathbf{P}(X\in\mathrm{d}x)\mathbf{P}(Y\in\mathrm{d}y) \tag{by change of var.} \\ &= \iint_{\mathbb{R}^2} f(x, y)^2 \, \mathbf{P}((X,Y)\in\mathrm{d}x\mathrm{d}y) \tag{by independence} \\ &= \mathbf{E}[f(X,Y)^2]. \end{align*}

More fundamentally, this identity holds because the distribution of $(\tau, \omega) \mapsto (X(\tau), Y(\omega))$ on $\Omega^2$ is the same as that of $(X, Y)$ on $\Omega$ by the independence of $X$ and $Y$.

$\endgroup$
3
  • $\begingroup$ First of all: Thanks a lot, that really helped! Especially your last point provides a nice view on independence, which I haven't realized before. $\endgroup$ May 21 at 16:22
  • $\begingroup$ And a quick follow-up question: We could generalize the result, so that in the same setting for measurable functions $g,h$ and $$g(a) = \mathbb{E}(h(X,a))$$ we obtain $$\mathbb{E}(g(Y)) = \mathbb{E}(h(X,Y)),$$ or could something go wrong there? $\endgroup$ May 21 at 16:24
  • $\begingroup$ @student7481 Of course the same argument will hold if $X$ and $Y$ are independent and the expectation of $h(X, Y)$ makes sense :). $\endgroup$ May 21 at 16:33

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.