binomial random variable Let $Y$ be a binomial random variable with $n$ trials and probability of success given by $p$. Show that $n-Y$ is a binomial random variable with $n$ trials and probability of success given by $1-p$.
I understand why this is true, and can explain in logically, but I just don't know how to prove it in mathematical terms.
Any help is greatly appreciated!
Thanks!
 A: A discrete random variable is determined by the probabilities of all possible outcomes. If $Y$ is $n$ trials of a binomial random variable with probability $p$ of success, then the probability that $Y = i$ is given by ${\displaystyle{n \choose i} p^i(1 - p)^{n-i}}$. Thus the probability that $n - Y = i$ is given by the probability that $Y = n - i$, or ${\displaystyle {n \choose  n - i} p^{n-i}(1 - p)^i}$. Since ${\displaystyle{n \choose n - i} = {n \choose i}}$, this is the same as ${\displaystyle{n \choose  i} (1 - p)^ip^{n-i}}$. But if $Z$ denotes $n$ trials of a binomial random variable with probability $1 - p$ of success, this is exactly the probability that $Z = i$. Thus $n - Y$ and $Z$ have the same distribution (i.e. the same probability of each outcome), which is equivalent to saying that $n - Y$ is also $n$ trials of a binomial random variable with probability $1 - p$ of success.
A: I am not really sure how I qould go about giving a full prove right now, but here is a suggestion. Rewrite $Y=\Sigma_{i=1}^n Y_i$ where the $Y_i$ are from your other question, they are the indicator functions for each trial. 
A: $Y$ counts the number of successes in $n$ trials. $n - Y$ counts the number of failures in $n$ trials. Notion of success and failure is arbitrary. $P[Y = i] = P[n - Y = n - i]$, so $n-Y$ is binomially distributed with "success" probability $1 - p$.
