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let and $ X_1\ and \ \ X_2$ be independent such that $ X_1 $ ~ binomial (m,1/2) and $ X_2 $ ~ binomial (n,1/2) , n $ \ne$ m

then how to prove that $ X_1 - X_2 + m $ ~ binomial (m + n ,1\2)

can a binomial variate be negative ? i mean let y = $ X_1 - X_2 + m $ , in case x1 =0 and x2 = n , then y= m-n , which could be negative if n > m ?

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  • $\begingroup$ No, binomial distribution's support is non-negative natural numbers. Btw, see math.stackexchange.com/questions/1065487/… $\endgroup$ – Tim Jan 11 '17 at 11:48
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    $\begingroup$ Replacing "$m$" by "$n$" would give $X_1 + (n-X_2)$. Since $n-X_2$ obviously is Binomial$(n,1-p)$ whenever $X_2$ is Binomial$(n,p)$ and the sum of independent Binomial$(,p)$ variables is Binomial$(,p)$, you would be done. $\endgroup$ – whuber Jan 11 '17 at 16:07
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(I'm assuming you have a typo and the constant in the random variable of interest should be $n$, not $m$. Otherwise the result cannot possibly hold, as already pointed out in the question.)

The moment-generating function of $T = X_1 - X_2 + n$ is equal to $$ M_T(t) = E[e^{(X_1 - X_2 + n)t}] = E[e^{X_1t}]E[e^{X_2(-t)}]e^{tn} = M_{X_1}(t)M_{X_2}(-t)e^{tn}, $$ where the second equality holds by the independence of $X_1$ and $X_2$. The moment-generating function of a $Bin(n, \theta)$-distributed random variable is $(1 - \theta + \theta e^t)^n$, see Wikipedia, and we have $$ M_T(t) = (0.5 + 0.5 e^t)^m (0.5 + 0.5 e^{-t})^n (e^t)^n = (0.5 + 0.5 e^t)^m (0.5 e^t+ 0.5)^n, $$ which further simplifies to $(0.5 + 0.5 e^t)^{m+n}$, the moment-generating function of $Bin(m + n, 1/2)$-distribution. Thus $X_1 - X_2 + n \sim Bin(m + n, 1/2)$.

In the question it was assumed that $m \neq n$ but the above did not use that in any way and as such holds also if $m = n$.

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