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I need to show the following:

I have two binomial random variables $X \sim BIN(m,p_1)$ and $Y \sim BIN(m,p_2)$, where $p_2 \geq p_1$. I want to show for any fixed constant $c \in \{0,...,m-1\}$ that

$P(Y \geq c \mid Y < m) \geq P(X \geq c \mid X < m)$

This holds with equality when $c=0$ (easy to see), and also holds easily when the condition of being less than $m$ is removed. It makes intuitive sense: the variable with the larger success probability should have at least as high probability of taking on larger values. Nonetheless, I don't see a good way to prove this explicitly. Perhaps there is a clever way to do so.

Any help would be appreciated! Thank you!

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4 Answers 4

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Here is my attempt at a solution, it proves the inequality in the other direction for the probability of the complement event from how I originally stated the problem.

enter image description here

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The probabilities can be computed as

$$P(X \geq n \vert X < m) = \frac{P(n \leq X \leq m)}{P(X < n)+P(n \leq X \leq m)} $$

or

$$P(X \geq n \vert X < m)^{-1} = 1 + \frac{P(X < n)}{P(n \leq X \leq m)}$$

When we increase the value of $p$ then this term $$\frac{P(X < n)}{P(n \leq X \leq m)}$$ is monotonically decreasing. This is because the terms in the denominator and numerator change at different rates.

  • We can express the probability of the one as a function of the other

    $$\frac{P(Y=k)}{P(X=k)} = \left(\frac{p_2/(1-p_2)}{p_1/(1-p_1)} \right)^k = f(k)$$

    where this term $f(k)$ is a monotonic increasing function of $k$ if $p_2 > p_1$.

  • We can express related inequalities like

    $$\frac{P(Y < n)}{P(X < n)} \leq f(n)$$

    $$\frac{P(n \leq Y \leq m)}{P(n \leq X\leq m)} \geq f(n)$$

    or

    $$\frac{P(n \leq X \leq m)}{P(n \leq Y\leq m)} \leq f(n)^{-1}$$

  • If we multiply the two, the following follows

    $$\frac{\left(\frac{P(Y < n)}{P(n \leq Y \leq m)}\right)}{\left(\frac{P(X < n)}{P(n \leq X \leq m)}\right)} \leq 1$$

    and

    $$\frac{\left(1+\frac{P(Y < n)}{P(n \leq Y \leq m)}\right)}{\left(1+\frac{P(X < n)}{P(n \leq X \leq m)}\right)} = \frac{P(X \geq n \vert X < m)}{P(Y \geq n \vert Y < m)} \leq 1$$

    from which follows

    $$P(Y \geq n \vert Y < m) \geq P(X \geq n \vert X < m)$$

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\begin{align*} P(Y\geq c| Y<m) - P(X \geq c|X<m) &= \sum_{t=c}^{m-1}{m \choose t}\big[p_2^t(1-p_2)^{m-t}-p_1^t(1-p_1)^{m-t}\big] \end{align*}

Consider $f(t) = p_2^t(1-p_2)^{m-t}$ and $g(t) = p_1^t(1-p_1)^{m-t}$

\begin{align*} h(t) &= \log(f(t))- \log(g(t))\\ &= t(\log(p_2)-\log(p_1)) +(m-t)(\log(1-p_2)-\log(1-p_1)) \end{align*}

\begin{align*} h'(t) &= \log(p_2)-\log(p_1) - (\log(1-p_2)-\log(1-p_1))\\ &= \log\big(\frac{p_2(1-p_1)}{p_1(1-p_2)}\big) \end{align*}

Given $\frac{p_2}{p_1} \geq 1$[$p_1 \neq 0$] $\implies \frac{1-p_1}{1-p_2}\geq 1$

Thus, $\frac{p_2(1-p_1)}{p_1(1-p_2)} \geq 1$

$\implies \log\big(\frac{p_2(1-p_1)}{p_1(1-p_2)}\big)$

$\implies h'(t) \geq 0 \implies \ h(t)$ is non-decreasing

$\implies h(t) \geq 0 \ \forall t \ \implies \frac{f(t)}{g(t)}\geq 1\ \implies P(Y\geq c| Y<m) - P(X \geq c|X<m) \geq 0$

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  • $\begingroup$ Thank you! A couple of things though, I think there needs to be a division to account for the condition, (i.e. divide f by $1-p_2^m$ and g by $1-p_1^m$, although since this is just a constant (WRT t) it doesn't alter the derivative of h. Also, I'm a little worried by the fact that c never comes into play. It seems as though you could follow this same line of reasoning to conclude that $P(Y<c \mid Y<m) - P(X<c \mid X<m) \geq 0$, just by replacing the bounds of the summation to go from 0 to c-1 instead of c to m-1. $\endgroup$
    – user103699
    Commented Feb 11, 2016 at 16:40
  • $\begingroup$ @PWM That fact is captured in the summation, so ideally instead of $\forall\ t$ it should be for $t \in [c,m-1]$, but that holds anyway $\endgroup$
    – Saket Choudhary
    Commented Feb 11, 2016 at 17:03
  • $\begingroup$ Still feeling a bit uneasy about this. I agree that $h(t)$ is non-decreasing, but I don't see how this implies that $h(t) \geq 0$ for all $t$. For instance if $p_2=0.5, p_1=0.25, m=2$ then $\frac{f(0)}{g(0)} = 4/9 \leq 1$ $\endgroup$
    – user103699
    Commented Feb 11, 2016 at 18:55
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    $\begingroup$ This does not work. The RHS of the first line is $$P(Y\ge c)-P(X\ge c),$$ not the desired $$P(Y\ge c\mid Y<m)-P(X\geqslant c\mid X<m).$$ $\endgroup$
    – Did
    Commented Feb 14, 2016 at 18:22
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Hint: Define a multinomial random vector $(Z_1, Z_2, Z_3)$ with parameters $m$ and $(p_1, p_2-p_1, 1-p_2)$. Then $X$ has the same distribution as $Z_1$ and $Y$ has the same distribution as $Z_1+Z_2$.

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    $\begingroup$ On second thought, although this is a way to prove the inequality without the conditions ($X<m$, $Y<m$), it's not clear that it will help in the case you want, in which these conditions are there. An alternative approach would be to use the binomial pmf to explicitly write out the conditional probability as a rational function of $p$, and then show that this function is increasing on the interval $(0,1)$, e.g. by showing the derivative is positive. $\endgroup$
    – Brent Kerby
    Commented Feb 11, 2016 at 0:03
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    $\begingroup$ Brent, think I have shown the result using your idea (using the derivative of the probability as a function of p) I plan on typing it up and will add it as an answer. $\endgroup$
    – user103699
    Commented Feb 11, 2016 at 20:09

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