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my teacher said that when you have two random variables X and Y, both are binomially distributed, then X-Y can never be binomail distribued? Why? I recall he mentioned something because it only takes on negative values??

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up vote 3 down vote accepted

This answer assumes independence of $X$ and $Y$.

A binomial distributed random variable $X$ with parameters $n$ and $p$ is a random variable with probability function given by $$ p_X(k)=P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\quad k=0,1,\ldots,n, $$ and in particular $p_X(k)=0$ for $k$ being a negative integer. So if you encounter a random variable $Z$ with the property that $P(Z=k)>0$ for some negative integer $k$, then $Z$ cannot be binomial distributed.

Now let $X\sim \mathrm{bi}(n,p)$ and $Y\sim \mathrm{bi}(m,q)$ be independent variables. This means that $$ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\quad k=0,1,\ldots,n $$ and $$ P(Y=k)=\binom{m}{k}q^k(1-q)^{m-k},\quad k=0,1,\ldots,m. $$ Let $Z=X-Y$, then $$ P(Z=-1)\geq P(X=0,\, Y=1)=P(X=0)P(Y=1)>0, $$ and hence $Z$ cannot be binomial distributed.

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Could it be "binomially distributed" with some offset? Would $X - Y + max(Y)$ be binomially distributed? – Joe Z. Jan 4 '13 at 14:57
@Joe Zeng: Try for example $X,Y\sim \mathrm{bi}(1,p)$ and put $q=1-p$. If $Z=X-Y+1$, then $P(Z=0)=pq$, $P(Z=1)=p^2+q^2$ and $P(Z=2)=qp$ which aren't binomial probabilities. – Stefan Hansen Jan 4 '13 at 16:19
You assume independence; that's not given as a hypothesis; see my answer for counterexamples in the dependent case. – joriki Jan 6 '13 at 9:55
@joriki: My bad, I read it as they were independent. Thanks for pointing it out. – Stefan Hansen Jan 6 '13 at 10:02

That statement is false. For $X=Y$, we have $X-Y=0$, so $X-Y$ is binomially distributed with parameter $n=0$. (Note that e.g. the Wikipedia article explicitly allows $n=0$.)

Also, if $X$ is binomially distributed with $n=1$ and $p=p_X$ and $Y$ is binomially distributed with $n=1$ and $p=p_Y\lt p_X$, and $\Pr(X=0,Y=1)=0$, then $X-Y$ is binomially distributed with $n=1$ and $p=p_X-p_Y$.

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