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Let $X$ and $Y$ be i.i.d binomial random variables with parameters $n$ and $\dfrac{1}{2}$ and let Z be another binomial random variable with parameter $2n$ and $\dfrac{1}{2}$. Then $P(X=Y)$ equals?

(a) $P(Z=0)$

(b) $P(Z=n)$

(c) $P(Z=2n-1)$

(d) $P(Z=n+1)$

I did this : $P(X=Y)$ --> $P(X-Y=0)$ then i thought $X$ and $Y$ follows same distribution subtracting them would give zero that is $X-Y$ $\sim$ $B(0,\dfrac{1}{2})$

But that doesnt seem to help me. Any ideas?

Daman
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2 Answers2

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Observe that $n-Y$ also has binomial distribution with parameters $n$ and $0.5$, and that $X$ and $n-Y$ are independent.

Then $S:=X+(n-Y)$ will have binomial distribution with parameters $2n$ and $0.5$.

Now observe that: $$X=Y\iff S=n$$

So: $$P(X=Y)=P(S=n)=P(Z=n)$$

drhab
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  • How did $n-Y$ follows binomial distribution ? – Daman Jan 08 '18 at 13:28
  • $P(n-Y=k)=P(Y=n-k)=\binom{n}{n-k}0.5^n=\binom{n}{k}0.5^n=P(Y=k)$. Also if $Y$ denotes the number of "successes" then $n-Y$ is the number of "failures" (hence binomially distributed with parameters $n$ and $1-p$. – drhab Jan 08 '18 at 13:30
  • Ohh is it like if n=30 . P(x=1)=p(x=29)? – Daman Jan 08 '18 at 13:34
  • In this case "yes" because the parameter is $p=0.5$ so that $p=1-p$. The number of successes follows the same distribution as the number of failures. – drhab Jan 08 '18 at 13:41
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Using independance of $X$ and $Y$, it comes \begin{align*} P(X=Y) & = \sum_{k=0}^n P((X=k)\cap(Y=k)) \\ & =\sum_{k=0}^n P(X=k)P(Y=k) \\ & =\frac{1}{2^{2n}}\sum_{k=0}^n \binom{n}{k}^2 \end{align*} It is well-known (at least a classical result) that $\sum_{k=0}^n \binom{n}{k}^2=\binom{2n}{n}$, so $$ P(X=Y)=\frac{1}{2^{2n}}\binom{2n}{n}=P(Z=n)$$

PS : your assertion that $X-Y\sim B(0,\frac{1}{2})$, makes no sense at all...

Netchaiev
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  • Like in iid random variables from normal distribution we subtract two random variables mean gets subtracted or can b added. I had that thing in my mind . Could you tell me where my confusions lies? – Daman Jan 08 '18 at 13:31
  • This well-known result can easily be proved by means of my answer. – drhab Jan 08 '18 at 13:39
  • @drhab okay. And about that classical result you mentioned can i have its proof somewhere? – Daman Jan 08 '18 at 13:42
  • @Daman In this answer it is shown that $P(X=Y)=2^{-2n}\sum_{k=0}^n\binom{n}{k}^2$. In my answer it is shown directly that $P(X=Y)=P(Z=n)=2^{-2n}\binom{2n}{n}$. This together proves that $\sum_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}$. – drhab Jan 08 '18 at 13:46
  • @Daman : you can add two independent binomial r.v., but you can't subtract them ; as drhab said, $X-Y$ can have negative value and, at least, is not binomial! But also as he showed, you can add $X$ and $n-Y$ ... it might remind you a little the addition in $\mathbb{Z}/n\mathbb{Z}$, if you are familiar with arithmetic. – Netchaiev Jan 08 '18 at 13:47
  • @Daman : it is proven here : https://math.stackexchange.com/questions/373122/sum-of-square-binomial-coefficients – Netchaiev Jan 08 '18 at 13:57