How to show a binomial random variable dominates another binomial random variable with a smaller success value?

Question

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!

Answer 1

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

Answer 2

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.