CDF of Binomial decreases with more trials. Let $X \sim Bin(n, \frac{c}{n})$, and $Y \sim Bin(n+1, \frac{c}{n+1})$. We know that $\mathbb{E}[X] = \mathbb{E}[Y] = c$. I am curious to know whether
$$
\Pr[Y \leq c] \leq \Pr[X \leq c].
$$
I tried confirming this numerically for some values of $c$ and $n$, and it seems to hold, which suggests it would be true in general. However, I have no idea how to prove this. Any help or a counterexample would be greatly appreciated!
 A: Notice that the median of the binomial distribution is either $\lfloor c\rfloor$ or $\lceil c\rceil$. Therefore if $c$ is an integer then
$$Pr\left[ Y\leq c\right]=Pr\left[X\leq c\right]=1/2$$
Now the non integer case.
This is also true if the median is $\lfloor c\rfloor$:
$$Pr\left[ Y\leq c\right]=Pr\left[Y\leq \lfloor c\rfloor\right]=1/2$$
and the same for $X$. What happens if the median is $\lceil c\rceil$? In that case we have that
$$Pr\left[ Y\leq c\right]=Pr\left[Y\leq \lfloor c\rfloor\right]=1/2-Pr\left[Y=\lceil c\rceil\right]$$
So all you need to do is compare $Pr\left[Y=\lceil c\rceil\right]$ with $Pr\left[X=\lceil c\rceil\right]$. I’m not sure that you can find a universal result for this comparison because some terms of $$Pr\left[Y=\lceil c\rceil\right]$$ are increasing in $n$ and others are decreasing. However if you look at the plot you can see that close to the mean the binomial converges to Poisson from above, so that would be a “visual proof” that your claim doesn’t hold. I am not sure about this last thing, though.
