I'm trying to solve the following exercise:
Suppose $X_1$ and $X_2$ are i.i.d. random variables with common $\mathcal{N}(0,1)$ normal distribution. Define $Y_n = X_1(\frac{1}{n} + |X_2|)^{-1}$. Use Fubini's theorem to verify that $\mathbb{E}(Y_n) = 0$. Note that as $n\to \infty$, $Y_n \to Y = X_1 |X_2|^{-1}$ and that the expectation of $Y$ does not exists, so this is one case where random variables converge but means do not.
I really don't know how to start solving this. Can you give me any hint (not the solution, but hints)? Thank you very much in advance
P.S.: This exercise appears is A Probability Path, by Resnick, chapter 5 (which is about integration and expectation).
P.P.S.: I had an idea, but I do not use Fubini's theorem: given that $X_1$ is independent of $(\frac{1}{n} + |X_2|)^{-1}$ (because the composition is measurable), and then $$ \mathbb{E}(Y_n) = \mathbb{E}(X_1) \mathbb{E} \left[ \left(\frac{1}{n} + |X_2| \right)^{-1} \right] = 0$$
You can start like this,
Since, $Y_n = X_1 [\frac{1}{n} + |X_2 |]^{-1}$ $$Pr[Y_n \leq y] = Pr[X_1 [\frac{1}{n} + |X_2 |]^{-1} \leq y] \\ = Pr[X_1 \leq y [\frac{1}{n} + |X_2 |]] \\ = \int_{x_2=-\infty}^{\infty} \int_{x_1=-\infty}^{y[\frac{1}{n} + |X_2 |]} \phi(x_1).\phi(x_2) dx_1 dx_2 \\ = \int_{x_2=-\infty}^{\infty} \phi(x_2) \int_{x_1=-\infty}^{y[\frac{1}{n} + |X_2 |]} \phi(x_1) dx_1dx_2 \\ = \int_{x_2=-\infty}^{\infty} \phi(x_2) . \Phi(y[\frac{1}{n} + |X_2 |) dx_2$$
Then find the pdf of $Y_n$ by differentiation of $Pr[Y_n \leq y]$. After that use the formula for finding expectation, $$E(Y_n) = \int_{y} y * \text{pdf of Y_n} dy$$ The range of $Y_n$ can be found from the second line.
NOTE$ \phi$ and $\Phi$ has been used to denote pdf and cdf