I have a binomial distribution, with Random Variable Y and n trials. r is an integer. How can I show that P(Y ≥ r) = P(Y ≤ n − r)? I think it involves using the binomial theorem. I have tried taking expressing P(Y ≥ r) as $$\sum_{x = 0}^{n} \binom{n}{x} p^x(1-p)^{n-x} - \sum_{x=0}^{r} \binom{n}{x} p^x(1-p)^{n-x}$$ and also: $$\sum_{x=r}^{n} \binom{n}{x} p^x(1-p)^{n-x} = \sum_{x=0}^{n-r} \binom{n}{x+r} p^{x+r}(1-p)^{n-(x+r)}$$ but I am unsure of how to proceed next. Can someone give me a hint?

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    $\begingroup$ +1 for the clear question and asking for a hint. If this is for homework or for self-study, you should add the self-study tag to your question (see stats.stackexchange.com/tags/self-study/info). $\endgroup$ – Patrick Coulombe Feb 5 '15 at 3:31
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    $\begingroup$ You need to change some $(p-1)$s to $(1-p)$. With questions like these it is often worth substituting some values in to try to understand why the equation works - if you tried a $p\neq0.5$ you'd see that actually it doesn't! $\endgroup$ – Silverfish Feb 5 '15 at 8:48

I think you'll find it difficult to prove, because it is not true unless $p = 0.5$.

Consider the simple case of $n = 1$, $r = 1$. Then:

$P(Y \geq r) = P(Y = 1) = p$

$P(Y \leq n - r) = P(Y = 0) = 1 - p$.

Starting with the LHS of your second equation, if you substitute $j = n - x$ and use ${n \choose j} = {n \choose {n-j}}$, you should be able to verify that

$P(Y \geq r) = \sum_{j=0}^{n-r} {n \choose j} (1-p)^j p^{n-j}$,

which is $P(X \leq n-r)$ if $X$ is a binomial random variable with success probability $(1-p)$.

This makes sense intuitively if you think of $Y$ as the number of successes and $X$ as the number of failures: if there are at least $r$ successes, there must be no more than $n-r$ failures.

  • $\begingroup$ I am still unclear on how you managed to change the upper and lower bounds of the sum? $\endgroup$ – roro172 Feb 5 '15 at 20:59
  • $\begingroup$ Counting upwards from $x=r$ to $x=n$ is equivalent to counting downwards from $j=n-r$ to $j=0$ ($j$ being the 'distance' between $x$ and $n$). You can just substitute $j = n - x$ into the limits of the sum: $\sum_{x=r}^n \equiv \sum_{j=n-r}^{n-n} \equiv \sum_{j=n-r}^0 \equiv \sum_{j=0}^{n-r}$ $\endgroup$ – Mark Feb 5 '15 at 21:40

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