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.


1 Answer 1


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 $Y$.

  • $\begingroup$ 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

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