Durrett Exercise 2.4.3 or 3.4.3 --- Symmetry of $Y_{k}\mathbb{1}_{|Y_{k}|\leq m}$ and $Y_{k}\mathbb{1}_{|Y_{k}|>m}$

Question

This is exercise from Durrett 2.4.3 or 3.4.3 depending on the version of the book. Part of this exercise asks to prove that

For $m>0$, let $U_{k}:=Y_{k}\mathbb{1}_{|Y_{k}|\leq m}$ and $V_{k}=Y_{k}\mathbb{1}_{|Y_{k}|>m}$. Show that for every $u<\infty$ and all $n$, we have $$\mathbb{P}\Big(\sum_{k=1}^{n}Y_{k}\geq u\sqrt{n}\Big)\geq\mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geq u\sqrt{n}, \sum_{k=1}^{n}V_{k}\geq 0\Big)\geq\dfrac{1}{2}\mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geq u\sqrt{n}\Big).$$

I have showed the first inequality using the fact that $Y_{k}=U_{k}+V_{k}$, but I don't know how to show the second one.

There have been a solution of this exercise but only said the second one is using symmetry. The same question was asked and answered here but I don't think the argument is correct (or precise) for the symmetric part. Does the convergence of $S_n / \sqrt{n}$ in distribution imply $EX_i = 0$?

Could somebody explain to me in detail what symmetric we are using here?

Answer 1

Here the random variables $Y_k$, $1\leqslant k\leqslant n$ have the same distribution and are symmetric in the sense that $Y_k$ has the same law as $-Y_k$.

The key point here is that

the vectors $(U_1,V_1,U_2,V_2,\dots,U_n,V_n)$ and $(U_1,-V_1,U_2,-V_2,\dots,U_n,-V_n)$ have the same distribution.

Indeed, since the vectors $(U_k,V_k)$, $1\leqslant k\leqslant n$ are independent, it suffices to show that $(U_1,V_1)$ has the same distribution as $(U_1,-V_1)$. The characteristic function of the second vector is given by $$ \varphi(s,t):=\mathbb E\left[\exp\left(i\left(sU_1-tV_1\right)\right)\right] = \mathbb E\left[\exp\left(i sY_1 \right)\mathbf 1_{\lvert Y_1\rvert\leqslant m}\right]+\mathbb E\left[\exp\left(-i tY_1 \right)\mathbf 1_{\lvert Y_1\rvert\gt m}\right]. $$ Using symmetry of $Y_1$ for the last expectation gives that $$ \varphi(s,t) = \mathbb E\left[\exp\left(i sY_1 \right)\mathbf 1_{\lvert Y_1\rvert\leqslant m}\right]+\mathbb E\left[\exp\left( i tY_1 \right)\mathbf 1_{\lvert Y_1\rvert\gt m}\right], $$ which is also the characteristic function of $(U_1,V_1)$.

Using the proven fact, one gets $$ 2\mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geqslant u\sqrt{n}, \sum_{k=1}^{n}V_{k}\geqslant 0\Big)= \mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geqslant u\sqrt{n}, \sum_{k=1}^{n}V_{k}\geqslant 0\Big)+\mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geqslant u\sqrt{n}, \sum_{k=1}^{n}V_{k}\leqslant 0\Big)=\underbrace{\mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geqslant u\sqrt{n}, \sum_{k=1}^{n}V_{k}= 0\Big)}_{\geqslant 0}+\mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geqslant u\sqrt{n}, \sum_{k=1}^{n}V_{k}\leqslant 0\Big)+\mathbb{P}\Big(\sum_{k=1}^{n}U_{k}\geqslant u\sqrt{n}, \sum_{k=1}^{n}V_{k}\gt 0\Big). $$