# Upper bound of sum of random variables

Suppose that we have random variables $$X_1$$, $$X_2$$ each drawn independently from Irwin Hall Distribution with same mean of 0 and different variances.

If I have $$B_1$$ and $$B_2$$, which are high probability upper bound of $$X_1$$ and $$X_2$$ respectively, can I presume the upper bound of $$X_1 + X_2$$?

In other words, If $$Pr[X_1 \gt B_1] \lt \epsilon$$, $$Pr[X_2 \gt B_2] \lt \epsilon$$ holds with negligible probability $$\epsilon$$, Can I find the tight upper bound $$B$$ s.t. $$Pr[X_1 + X_2 \gt B] \lt \epsilon$$?

I thought of two candidates for upper bound B. One is $$B_1 + B_2$$ and the other is $$\sqrt{B_1^2+B_2^2}$$.

I thought that first candidate must satisfy the upper bound of $$X_1 + X_2$$, But it would be a loose bound. Second one would be better and tight upper bound, but I can't prove $$Pr[X_1 + X_2 \gt \sqrt{B_1^2+B_2^2}] \lt \epsilon$$

I got idea from variance of sum of random variables. If $$X_1$$ ~ $$(0, \sigma_1^2)$$ , $$X_2$$ ~ $$(0, \sigma_2^2)$$, $$X_1 + X_2$$ will follow the distribution of $$(0, \sigma_1^2 + \sigma_2^2)$$ Is this thought reasonable?

More generally, I also want to know the upper bound of $$X_1$$, $$X_2$$, ... , $$X_n$$. If the second candidate is true, I would like to know how to prove it.

Thank you for reading.

• The distribution of the sum $Y = X_1+X_2$ is given by $P_Y(y) = \int P_{X_1}(x)P_{X_2}(y-x)dx.$ Your high probability upper bound then follows from solving $\int_{B}^\infty P_Y(y)dy < \epsilon.$ Suggest to set this equation up and try to get $B$ from it. Then you can generalize to the sum of $n$ variables Commented Aug 3, 2023 at 6:43

The event $$X_1+X_2 > B_1+B_2$$ is contained in the union of the two events $$X_1>B_1$$ and $$X_2>B_2$$. (This assertion is equivalent to the algebraic implication "if $$X_1+X_2>B_1+B_2$$ then $$X_1>B_1$$ or $$X_2>B_2$$", which is equivalent to the easily verified "if $$X_1\le B_1$$ and $$X_2\le B_2$$ then $$X_1+X_2\le B_1+B_2$$".)
In particular, if $$Pr[X_1>B_1]<\varepsilon$$ and $$Pr[X_2>B_2]<\varepsilon$$, then $$Pr[X_1+X_2>B_1+B_2] \le Pr[X_1>B_1] + Pr[X_2>B_2] < 2\varepsilon.$$ It's not hard to show by example that this bound is best possible: let $$X_j$$ take only values $$B_j-1$$ and $$B_j+2$$ with probabilities $$1-\delta$$ and $$\delta$$ where $$\delta$$ is just less than $$\varepsilon$$, anticorrelated so that $$X_1=B_1+2$$ and $$X_2=B_2+2$$ never occur simultaneously.