# Cumulative distribution of Binomial Random variable

How to prove this:

Proposition. If $F_n$ is the distribution function of a $\textrm{Bin}(n,p)$ random variable, then, for every real fixed $t$, the sequence $\{F_n(t)\}_{n=1}^\infty$ is nonincreasing.

P.S. There was some heavy editing here.

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I don't follow your argument at all. Isn't the question whether or not when n$_1$>n$_2$ P{X$_1$<=k)<P{X$_2$<=k} for each k<=n$_2$? I don't even think this is true. It could depend on p. –  Michael Chernick Sep 7 '12 at 0:39

Suppose that you have a sequence $$Y_1,Y_2,Y_3,\dots$$ of independent and identically distributed random variables such that $Y_1\sim\textrm{Bernoulli}(p)$.

Define $F_n(t)=P\left\{\sum_{i=1}^n Y_i\leq t\right\}$, which is the distribution function of a $\textrm{Bin}(n,p)$ random variable.

If $\sum_{i=1}^n Y_i>t$, then $\sum_{i=1}^{n+1} Y_i>t$. Hence, we have the inclusion $$\left\{\omega:\sum_{i=1}^n Y_i(\omega)>t\right\} \subset \left\{\omega:\sum_{i=1}^{n+1} Y_i(\omega)>t\right\} \, ,$$ and it follows from the monotonicity of $P$ that $$P\left\{\omega:\sum_{i=1}^n Y_i(\omega)>t\right\} \leq P\left\{\omega:\sum_{i=1}^{n+1} Y_i(\omega)>t\right\} \, ,$$ which is equivalent to $$1-F_n(t)\leq 1-F_{n+1}(t) \, ,$$ and so $$F_n(t)\geq F_{n+1}(t) \, ,$$ yielding that, for each fixed $t$, the sequence of real numbers $\{a_n\}_{n=1}^\infty$ defined by $a_n=F_n(t)$ is nonincreasing.

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It this is what the OP intended, may I edit his question to reflect that? Sorry for the little JMRT here. –  Zen Sep 7 '12 at 1:49
Thanks, Zen. This is what I was thinking. Please edit the questions as needed. –  user13154 Sep 7 '12 at 1:53
Is it "legal" for me to edit your question? –  Zen Sep 7 '12 at 1:56
(+1) Hopefully the OP will notice that the binomial distribution enters in only in the most nominal way in the proof. Indeed, this is true for any family of distributions formed from sums of exchangeable nonnegative random variables. –  cardinal Sep 7 '12 at 2:42
You are welcome, @user131154! If you like it, you may consider checking the answer as such. –  Zen Sep 7 '12 at 16:51
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