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If $X $~ $Bin(n, p)$ and $Y$ ~ $Bin(m, p)$ I want to show that $X-Y$ is not binomial.

I tried to use the technique of this proof given in lecture that Sum of two independent binomial variables is binomial with $X+Y$ ~ $Bin(n+m, p)$ analogous for the situation $X-Y$.

I suspect that $X-Y$ should be ~ $Bin(n-m, p)$, and therefore not binomial for the case that $m$ > $n$ as that would a negative amount of independent Bernoulli trials with a positive probability p; by definition of the Binomial distribution this makes no sense. Probability of "success" in zero or less Bernoulli trials should be zero, not positive

This would seem to agree with the solution given by the book:

"A Binomial can’t be negative, but X −Y is negative with positive probability." But is $X-Y$ a negative, or is it only negative for $m > n$?

Is this the right train of thought?

If so, is there a mathematical way to express $X-Y$ ~ $Bin(n-m, p)$?

user556576
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  • See here: https://math.stackexchange.com/questions/562119/difference-of-two-binomial-random-variables – Math1000 Apr 27 '18 at 05:21

2 Answers2

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No, the difference does not follow a binomial distribution, even when $m\le n$. The argument in the book is simply that $X-Y$ can be negative while a binomial cannot, and that suffices to prove the claim. Even when $m\le n,$ you can still have $X=0$ and $Y=1$ (Except in the trivial case where $m=0.$)

However, if you are in the regime where a binomial can be approximated by a normal, then the difference between two approximately normal things will be approximately normal (but that's beside the point here).

spaceisdarkgreen
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  • Thank your for the reply. So for example if X = 1 and Y = 0 then X-Y = 1. Although this is positive, X-Y is still not necessarily binomial because of the existence of a counterexample i.e. X = 0 and Y = 1 which is a contradiction of a binomial being positive? – user556576 Apr 27 '18 at 03:30
  • @TYBG . Yes, the key is that it is possible for $X-Y$ to be negative, (even if in some cases it might be very unlikely, like if $np$ is very large and $mp$ is very small). But a binomial RV can never be negative. (Again, this is ignoring corner cases like $m=0,$ $p=0$ or $p=1$.) – spaceisdarkgreen Apr 27 '18 at 03:33
  • Would it be correct to think of the Binomially distributed R.V.s $X$ and $Y$ as $n$ and $m$ independent Bernoulli trials and $X$ and $Y$ as the # of successes in those trials, with a shared probability parameter $p$ for $0 < p <1$, thus making it nonsensical for $X - Y$ to be Binomial (with a positive probability) when $Y > X$ (there are negative successes). – user556576 Apr 27 '18 at 03:42
  • @TYBG Some ways yes and some ways no. I think you're making a conceptual error when you say "binomial with positive probability". A distribution is either binomial or it isn't... "binomial" describes the range of possible outcomes and their probability. However it is very sensible to think of $X$ and $Y$ as representing the number of successes in a set of $n$ and another independent set $m$ trials. In fact this is an easy way to see $X+Y$ is binomial(m+n,p), since it is number of successes in $n+m$ trials. – spaceisdarkgreen Apr 27 '18 at 03:48
  • @TYBG On the other hand $X-Y$ is "the difference between the number of successes in $n$ independent trials and the number of successes in an independent set of $m$ independent trials" and has no such simple characterization as "the number of successes in _ independent trials." – spaceisdarkgreen Apr 27 '18 at 03:49
  • My conceptual understanding is admittedly shaky. Will need to get my nose back into the textbook. It is my understanding that since $X-Y$ can't be characterized as "the number of successes in _ independent trials", which is an intuitive way to understand a binomial distribution, then it should follow that $X-Y$ isn't binomial. – user556576 Apr 27 '18 at 04:00
  • @TYBG Yep that's right, and the fact that it can be negative with some positive probability while a binomial can't proves beyond all doubt that there's no such interpretation possible. (So the part about "negative successes" not making sense in your previous comment was also on the mark). – spaceisdarkgreen Apr 27 '18 at 04:10
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$X+Y$ is binomially distributed since the sum of counts of successes in two independent sequences of $n$ and $m$ iid Bernoulli trials, each with success rates $p$, is equivalently stated as being the count of successes in a sequence of $n+m$ iid Bernoulli trials each with success rate $p$.

$X-Y$ is not binomially distributed because the difference between counts of successes in two independent sequences of $n$ and $m$ Bernoulli trials , each with iid success rates $p$, just cannot be expressed as being a count of successes in a sequence of some amount of iid Bernoulli trials.

Also as noted, when $X$ realises values less than $m$ it is possible for $X-Y$ to be negative, which is not the case for any Binomially Distributed random variable .

Graham Kemp
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  • Yes, your last paragraph is how I understood it and is based on the conditions of the binomial distribution. The sentence "cannot be expressed as being a count of successes in a sequence of some amount of iid Bernoulli trials" is helpful as a more intuitive way of understanding why the difference of the R.V.'s cannot be binomial. – user556576 Apr 27 '18 at 04:09