4
$\begingroup$

Let $X= x_1 + x_2 + \ldots + x_m$, $Y=y_1 + y_2 + y_3 + \ldots + y_n$, and $Y' = y'_1 + y'_2 + \ldots + y'_n$, where

  • Each $x_i$ is a Bernoulli variable which takes value $1$ with probability $p_i>0$.
  • Each $y_i$ is a Bernoulli variable which takes value $1$ with probability $q_i>0$.
  • For $i>2$, each $y'_i$ is a Bernoulli variable which takes value $1$ with probability $q_i>0$.
  • $y'_1$ and $y'_2$ are two Bernoulli variables which take value $1$ with probability $(q_1 + q_2)/2$ (Suppose that $q_1 \neq q_2$).
  • All the variables are independent.

Furthermore, suppose that $E(X) \geq E(Y)$ and $m>n$. Now, let $A = Pr(Y\geq X)$ and $A' = Pr(Y' \geq X)$. The question is whether $A>A'$ or $A'>A$?

$\endgroup$
3
  • $\begingroup$ Are you assuming $x_i, y_i, y'_i$ for i=1,..,n (i.e., all $3n$ r.v.'s) are all independent? $\endgroup$ Aug 4, 2018 at 15:00
  • $\begingroup$ Yes, all $x_i,y_i$, and $y'_i$'s are independent. $\endgroup$
    – Masood
    Aug 4, 2018 at 15:03
  • 1
    $\begingroup$ Please add the independence assumption in the statement of the problem. If $q_1=q_2$, then $A=A'$, so you would need an extra assumption to exclude the possibility. Also, I presume you want $p_i>0$ for otherwise, the assumption $m>n$ is pointless. $\endgroup$
    – Algernon
    Aug 4, 2018 at 20:14

1 Answer 1

4
$\begingroup$

Both $A>A'$ and $A'>A$ are possible.

Let $S:= Y_3+Y_4+\cdots Y_n$. Assuming independence, we have \begin{align} \triangle A := A-A' &= \mathbb{P}(Y\geq X) - \mathbb{P}(Y'\geq X) \\ &= \sum_a\mathbb{P}(X=a)\big[\mathbb{P}(Y\geq a) - \mathbb{P}(Y'\geq a)\big] \\ &= \sum_a\mathbb{P}(X=a)\Big[\sum_{b\geq a}\mathbb{P}(Y=b)-\sum_{b\geq a}\mathbb{P}(Y'=b)\Big] \\ &= \sum_a\mathbb{P}(X=a)\sum_{b\geq a}\sum_{c=0}^2\mathbb{P}(S=b-c) \big[\mathbb{P}(Y_1+Y_2=c) - \mathbb{P}(Y'_1+Y'_2=c)\big] \;. \end{align} Let $\bar{q}:=(q_1+q_2)/2$ and $\varepsilon:=|q_1-q_2|/2$. We can verify that \begin{align} \mathbb{P}(Y_1+Y_2=c) - \mathbb{P}(Y'_1+Y'_2=c) &= \begin{cases} -\varepsilon^2 &\text{if $c=0$,} \\ 2\varepsilon^2 &\text{if $c=1$,} \\ -\varepsilon^2 &\text{if $c=2$.} \end{cases} \end{align} It follows that \begin{align} \triangle A &= \sum_a\mathbb{P}(X=a)\sum_{b\geq a} \varepsilon^2\big[ -\mathbb{P}(S=b) + 2\mathbb{P}(S=b-1) - \mathbb{P}(S=b-2) \big] \\ &= \varepsilon^2\sum_a\mathbb{P}(X=a)\Big[ -\sum_{b\geq a}\mathbb{P}(S=b) +2\sum_{b\geq a-1}\mathbb{P}(S=b) -\sum_{b\geq a-2}\mathbb{P}(S=b) \Big]\\ &= \varepsilon^2\sum_a\mathbb{P}(X=a)\big[ \mathbb{P}(S=a-1)-\mathbb{P}(S=a-2) \big] \;. \end{align}

Now, let us consider two examples. Observe that if $n=2$, then $\mathbb{P}(S=0)=1$, and consequently \begin{align} \triangle A &= \varepsilon^2\big[\mathbb{P}(X=1)-\mathbb{P}(X=2)\big] \tag{$\star$} \end{align} Also, note that $\mathbb{E}(X)\geq \mathbb{E}(Y)$ if and only if $p_1+p_2+\cdots+p_m\geq q_1+q_2+\cdots+q_n$.

Example 1 ($A<A'$).
Let $m=4$, $n=2$, $p_1=p_2=p_3=p_4=1/2$, and let $0<q_1<q_2<1$ Then, $p_1+p_2+p_3+p_4=2>q_1+q_2$, hence the requirements are satisfied. From ($\star$) we get \begin{align} \triangle A &= \varepsilon^2(1/4)^4\Big[ \binom{4}{1}-\binom{4}{2} \Big] < 0 \;. \end{align}

Example 2 ($A>A'$).
Let $m=3$, $n=2$, $p_1=1-\delta$, $p_2=p_3=\delta$ where $0<\delta\ll 1$, and assume $q_1,q_2$ are distinct and satisfy $q_1+q_2\leq 1$. Then, $p_1+p_2+p_3=1+\delta>1\geq q_1+q_2$, hence the requirements are satisfied. From ($\star$) we get \begin{align} \triangle A &= \varepsilon^2\Big[ (1-\delta)^3+2\delta^2(1-\delta) - 2\delta(1-\delta)^2-\delta^3 \Big] \;. \end{align} which is positive (it is $\approx \varepsilon^2$) when $\delta>0$ is sufficiently small.

$\endgroup$
2
  • $\begingroup$ Thanks. It seems that most of the times we have $\Delta A$ is negative. Is there a sufficient condition that guarantees $\Delta A <0$? $\endgroup$
    – Masood
    Aug 5, 2018 at 11:35
  • $\begingroup$ I tend to agree with you, but I am afraid the intuition could come from the asymptotic and may not apply to non-asymptotic highly asymmetric cases as in Example 1 above. If $m$ and $n$ are large and $p_i,q_i$ are not wildly different from one another, the sums $S$ and $X$ are almost Gaussian, in particular, symmetric around their means. On the other hand, the distribution of $Y'_1+Y'_2+S$ is slightly more spread than the distribution of $Y_1+Y_2+S$, which together with the assumption $\mathbb{E}(Y)<\mathbb{E}(X)$, suggests that $Y'$ has more chance of being larger than $X$ than $Y$ does. $\endgroup$
    – Algernon
    Aug 5, 2018 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.