Does $E[X]\gg E[Y]$ for independent RV imply that $Pr[X+Y \geq x] \sim Pr[ X \geq x]$?

Question

We have two independent random variables $X$ and $Y$, where we know that $E[X]\gg E[Y]$, thus $\frac{E[Y]}{E[X]}\rightarrow 0$.

I am now interested in $Pr[X+Y \geq x]$ and would like to show that $$Pr[X+Y \geq x] \sim Pr[X \geq x]$$ for $x=O(1)$.

$X$ and $Y$ are both sums of independent Bernoulli random variables, thus $X=\sum_i X_i$ and $Y=\sum_j Y_j$ where $X_i, Y_j$ are independent.

I tried to use the first moment method on $\frac{X}{Y}$ and also tried to extract the case where $Y=0$ but this all does not help. Would really appreciate if you could give me a hint, as small as it may be. To write $$Pr[X+Y \geq x] = Pr[X \geq x] Pr[Y=0] + Pr[X=0] Pr[Y \geq x]+....$$ was my idea, but i don't see how this should help me.

Answer 1

No.

Let $X=10$ be a constant and $Y$ take two values, $100$ and $-100$, with equal probabilities. Then $E(X)=10\gg 0=E(Y)$, but $P(X>15)=0$ while $P(X+Y)>15=1/2$.

I don't think you can get a "universal" inequality for all $x$. Take $x$ such that $P(X>x)$ is small but $P(Y>x)$ is large. This does not contradict your condition on the expectations as is easily seen from the two examples you got.

What you would probably need is something like $P(X\gg Y)\approx 1$.

Note that the probability of density of $X+Y$ is the Convolution of their individual probability densities (which is a more complicated formula that the one you are trying to use).

Answer 2

A counter example with non-negative random variables would be when $Y$ is identically $1$ and $X$ is $0$ with probability $0.9$ and $10^{10}$ with probability $0.1$. Then $P(X + Y > x) = 1$ when $x < 1$, but $P(X > x) = 0.1$.