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Let $X\sim B(n,p_h)$ and $Y\sim B(n,p_\ell)$ be two random variables following a respective binomial distribution, where $p_h>p_\ell$. I want to show that $$P(X\ge\alpha)\ge P(Y\ge\alpha),$$ for any $\alpha\in\{0,1,\dots,n\}$. In other words, I want to show $$P(X\ge\alpha)=\sum_{i=\alpha}^{n}\binom{n}{i}p_h^i(1-p_h)^{n-i}\ge P(Y\ge\alpha)=\sum_{i=\alpha}^{n}\binom{n}{i}p_\ell^i(1-p_\ell)^{n-i}.$$

The statement to be proven is obviously true, but proving it is extremely difficult (at least for me). Can anyone give me some guidance on proving it?

It would be nice if someone can give an algebraic proof too, that is, show that $$\sum_{i=\alpha}^{n}\binom{n}{i}p_h^i(1-p_h)^{n-i}\ge \sum_{i=\alpha}^{n}\binom{n}{i}p_\ell^i(1-p_\ell)^{n-i}.$$

Thank you very much!

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    I wonder if it is possible to prove this algebraically, that is, show that $$\sum_{i=\alpha}^{n}\binom{n}{i}p_h^i(1-p_h)^{n-i}\ge \sum_{i=\alpha}^{n}\binom{n}{i}p_\ell^i(1-p_\ell)^{n-i}.$$ – 2017-02-11

2 Answers 2

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For $i=1,\dots n$ let $X_i$ have Bernouilli($p_h$) distribution.

For $i=1,\dots n$ let $U_i$ have Bernouilli($\frac{p_l}{p_h}$) distribution.

Also let it be that there is independency among these random variables.

For convenience let it be that $X_i(\omega),U_i(\omega)\in\{0,1\}$ for every $\omega\in \Omega$.

Now define $Y_i:=X_iU_i$ so that $Y_i\leq X_i$.

Then $Y_i$ has Bernouilli($p_l$) distribution.

If $X:=X_1+\cdots+X_n$ and $Y:=Y_1+\cdots+Y_n$ then they have the binomial distributions mentioned in your question.

Now observe that $X\geq Y$ so that $\{Y\geq\alpha\}\subseteq\{X\geq\alpha\}$ for any $\alpha$.

Consequently $\Pr(X\geq\alpha)\geq\Pr(Y\geq\alpha)$.

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    +1. For reference, this construction is called a coupling (between the two binomial distributions). – 2017-02-11
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    Thank you very much! No wonder I cannot crack the proof. I just wonder what do we mean by $Y_i\le X_i$ and $Y\ge \alpha$? For example, is it true that $Y_i\le X_i$ means $P(Y_i\le X_i)=1$? – 2017-02-11
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    Thank you very much for the reference, Did! It helps a lot! – 2017-02-11
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    $Y_i\leq X_i$ can be read as $Y_i(\omega)\leq X_i(\omega)$ for each $\omega\in\Omega$ or equivalently as $\{Y_i\leq X_i\}=\Omega$. It is a bit stronger than $Y_i\leq X_i$ a.s. It is not an essential thing in my answer, but more a choice for convenience. I repaired the statement that includes $Y\geq\alpha$ – 2017-02-11
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    I see. I am truly grateful for your explanation! This proof has bothered me for quite a while! – 2017-02-11
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    Glad to help. You are welcome. – 2017-02-11
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Let us consider the probability space $\Omega=[0,\,1]^n$ with $\sigma$-algebra of Borel subsets and Lebesgue measure as a probability measure.

Construct $2n$ Bernoulli r.v. $X_1,\ldots, X_n$ and $Y_1,\ldots, Y_n$ on this space such that they are independent within each vector, $\sum X_i \sim B(n, p_h)$, $\sum Y_i \sim B(n, p_l)$, and $X_i\geqslant Y_i$ for all $i=1,\ldots,n$.

Set for each $\vec \omega\in [0,\,1]^n$

$$X_i(\vec \omega)=\begin{cases}1, \ 0\leq \omega_i\leq p_h\cr 0, \ p_h < \omega_i\leq 1,\end{cases}$$ $$Y_i(\vec \omega)=\begin{cases}1, \ 0\leq \omega_i\leq p_l\cr 0, \ p_l < \omega_i\leq 1.\end{cases}$$

Then $Y_i\leq X_i$, $\sum Y_i \leq \sum X_i$. And therefore $$P(Y\geq a) = P\left(\sum Y_i\geq a\right) \leq P\left(\sum X_i\geq a\right)=P(X\geq a).$$

Other words, we construct each independent pairs $(X_i, Y_i)$ with Bernoulli distributions s.t. $\{Y_i=1\}$ imply $\{X_i=1\}$, but not vice versa.

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    Thank you very much for the proof! I really appreciate it! – 2017-02-11