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Let $X$, $Y$, $Z$, $A$ be a set of random variables drawn from the Irwin-Hall distribution where $X$ is the sum of $c$ iid r.v.s, $Y$ is the sum of $c$ iid r.v., $Z$ is the sum of $n - c$ iid r.v.s, and $A$ is the sum of $n - c$ iid r.v.s.

I want to compare $\Pr[(X \leq x) \cap (Z \leq x - X) \cap (A \leq x - X)]$ with $\Pr[(X \leq x) \cap (Y \leq x) \cap (Z \leq x - X) \cap (A \leq x - Y)]$. Intuitively, it seems like $\Pr[(X \leq x) \cap (Z \leq x - X) \cap (A \leq x - X)] \geq \Pr[(X \leq x) \cap (Y \leq x) \cap (Z \leq x - X) \cap (A \leq x - Y)]$ but I couldn't find a clean proof for this.

You can assume that $x \in [0, n]$.

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$\newcommand{\R}{\mathbb{R}}$ Welcome to MathOverflow!

Your conditions on $X$, $Y$, $Z$, $A$, as I understood them, imply that $X$, $Y$, $Z$, $A$ are independent nonnegative random variables (r.v.'s), with $(X,Z)$ equal $(Y,A)$ in distribution -- which is all we need to verify your conjecture.

Indeed, since $Z,A\ge0$, we see that event $\{Z \le x - X\}$ implies $\{X\le x\}$, and $\{A \le x - Y\}$ implies $\{Y\le x\}$. So, denoting by $F$ the common cdf of the iid r.v.'s $Z$ and $A$, we see that your conjecture simplifies to $p_1\ge p_2$, where \begin{equation} p_1:=P(X+Z\le x,X+A\le x)=\int_\R P(X\in du)P(Z\le x-u,A\le x-u) =\int_\R P(X\in du)F(x-u)^2=EF(x-X)^2, \end{equation} \begin{equation} p_2:=P(X+Z\le x,Y+A\le x)=P(X+Z\le x)^2= \Big(\int_\R P(X\in du)P(Z\le x-u)\Big)^2 =\big(EF(x-X)\big)^2. \end{equation} So, by the Cauchy--Schwarz inequality, we indeed have $p_1\ge p_2$.

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  • $\begingroup$ Thanks for the response, it's very nice! I have one more question. Would it be correct to say $\Pr[(X \leq x) \cap (Z \leq x - X) \cap (A \leq x - X)] = \Pr[(X \leq x) \cap (Y \leq x) \cap (Z \leq x - X) \cap (A \leq x - Y) | X = Y]$ and the inequality follows directly from that? $\endgroup$
    – user1246462
    Commented Jul 18, 2019 at 14:26
  • $\begingroup$ @user1246462 : I am glad you liked this answer. Concerning the question in your comment: You should clarify what you mean by conditioning on the zero-probability event $\{X=Y\}$. Anyhow, in general, conditioning does not preserve the probability. $\endgroup$
    – Iosif Pinelis
    Commented Jul 18, 2019 at 17:08
  • $\begingroup$ ah yes, I'd forgotten that the variables are continuous, not discrete, so the event $X = Y$ has zero-probability of happening. If all the variables are discrete, then we can show $P(X + Z \leq x, X + A \leq x) \geq P(X + Z \leq x, Y + A \leq x)$ by using the fact that $P(X + Z \leq x, X + A \leq x) = P(X + Z \leq x, Y + A \leq x | X = Y)$. $\endgroup$
    – user1246462
    Commented Jul 18, 2019 at 17:39
  • $\begingroup$ @user1246462 : The equality $P(X+Z\le x,X+A\le x)=P(X+Z\le x,Y+A\le x|X=Y)$ will not hold in general even if $X$ and $Y$ are discrete. Indeed, $P(X+Z\le x,X+A\le x)=\sum_uf(u)F(u)^2$, where $f$ is the pmf of $X$ and $F$ is the cdf of $Z$, whereas $P(X+Z\le x,Y+A\le x|X=Y)=P(X+Z\le x,X+A\le x|X=Y)=\sum_uf(u)^2F(u)^2/\sum_uf(u)^2$ $\endgroup$
    – Iosif Pinelis
    Commented Jul 18, 2019 at 19:12

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