Weighted sum of i.i.d. random variables

Question

Suppose we have a sequence of independent identically distributed random variables $X_1,\cdots,X_n$ with mean zero ($\mathbb{E} X_1=0$), and a sequence of positive weights $a_1,\cdots,a_n\geq 0$. Now we denote $M=\max_{1\leq i\leq n}a_i$, do we have that $$P\left(\left|\sum_{i=1}^na_iX_i\right|\geq t\right)\leq P\left(M\cdot\left|\sum_{i=1}^nX_i\right|\geq t\right),$$ and $$\mathbb{E}\left|\sum_{i=1}^na_iX_i\right|\leq M\cdot\mathbb{E}\left|\sum_{i=1}^nX_i\right|.$$
It is intuitively plausible that larger weights increase its variance, thus the tail probability and expectation should be enlarged. But I have not found a strict proof of this problem, or if someone can give a counter-example?

Answer 1

The inequality is not true.

For example take $X_{1},X_{2}$ to be Symmetric Bernoulli $\frac{1}{2}$ variates and let $a_{1}=3,a_{2}=2$ and let $t=1$

Then $P(|3X_{1}+2X_{2}|\geq 1)= 1$ as for any $X_{1}=i,X_{2}=j\,,i,j\in\{-1,1\}$ you have $|3i+2j|\geq 1$

And $P(3|X_{1}+X_{2}|\geq 1)=P(X_{1}=1,X_{2}=1)+P(X_{1}=-1,X_{2}=-1)=\frac{1}{2}$

What you do have is that by triangle inequality $|\sum_{k=1}^{n}a_{k}X_{k}|\leq \sum_{k=1}^{n}|a_{k}X_{k}|\leq M\sum_{k=1}^{n}|X_{k}|$

Thus by simple set inclusion $\{|\sum_{k=1}^{n}a_{k}X_{k}|\geq t\} \subseteq \{M\sum_{k=1}^{n}|X_{k}|\geq t\}$

Thus $P(|\sum_{k=1}^{n}a_{k}X_{k}|\geq t)\leq P(M\sum_{k=1}^{n}|X_{k}|\geq t)$

Now for any positive random variable $Y$, by Fubini's Theorem $E(Y)=\int_{0}^{\infty}P(Y\geq t)\,dt$

Thus you would also have $E(|\sum_{k=1}^{n}a_{k}X_{k}|)\leq E(M\sum_{k=1}^{n}|X_{k}|\geq t)$