Let $\{X_n, U_n\ |\ n \geq 1 \}$ be a collection of independent random variables such that $X_n$'s are iid random variables following standard Cauchy distribution and $U_n$'s are independent random variables following Bernoulli distribution with parameter $\frac 1 n.$ Define a sequence of random variables $\{Y_n \}_{n \geq 1}$ by $$Y_n = \begin{cases} 0 & \text{if}\ U_n = 0 \\ \frac {X_n} {n} & \text{if}\ U_n = 1 \end{cases}$$
Questions $:$
$(1)$ Does $Y_n \xrightarrow{L^1} 0\ $?
$(2)$ If $(1)$ is false, can we slightly modify the above hypothesis (without altering the distributions) so that $(1)$ holds under the modified hypothesis?
My Attempt $:$ $(1)$ is false because of the following reason $:$ $$\begin{align*} \mathbb E \left [|Y_n| \right ] & = \mathbb E \left [|Y_n|\ \lvert\ U_n = 0 \right ] \mathbb P (U_n = 0) + \mathbb E \left [|Y_n|\ \lvert\ U_n = 1 \right ] \mathbb P (U_n = 1) \\ & = \frac {\mathbb E \left [|X_n| \right ]} {n^2} \\ & = \infty\end{align*}$$ as $\mathbb E [|X_n|] = \infty$ for all $n \geq 1.$ But what about $(2)\ $? Do anyone have any idea about it?
Regards
Anil
I assume in your second equation you actually meant $$ \frac {\mathbb E \left [|X_n| \right ]} {n^2}$$ As long as you fix the distributions and keep independence you always get this result. You also don't need conditional expectations to show.
In your setting you have $$Y_n = \frac{1}{n}\cdot X_n\cdot 1_{\{U_n = 1\}}$$ and so \begin{align*}E[|Y_n|] &= E\left[\frac{1}{n}\cdot |X_n|\cdot 1_{\{U_n = 1\}}\right] \\ &= \frac{1}{n}\cdot E[|X_n|]\cdot P(U_n = 1) \end{align*}
by independence.
Also when you take an arbitrary generalized Bernoulli distribution for $U_n$ with value $a_n > 0$ instead of $\frac{1}{n}$ and probability $p_n$ the result above would just change to $$E[|Y_n|] = a_n \cdot E[|X_n|]\cdot p_n$$ but still this equals $+\infty$ due to $$E[|X_n| = +\infty$$ because $X_n$ is Cauchy.
So, Cauchy $X_n$, Bernoulli $U_n$ and independence leads to your result.