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jamcc09

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As you may notice from a previous post of mine, I am working through the Blitzstein & Hwang probability book. I have run into a problem, which I cannot quite solve. It goes like this:

Alice flips a fair coin n times and Bob flips another fair coin n + 1 times, resulting in independent $X \sim Bin(n, 1/2)$ and $Y \sim Bin(n+1, 1/2)$.

(a) Let $V = min(X,Y)$ and $W = max(X,Y)$. Find $E(V) + E(W)$. If $X=Y$ then $X=Y=V=W$.

(b) Show that $P(X<Y) = P(n-X<n+1-Y)$.

(c) Compute $P(X<Y)$. Hint: Use (b) and the fact that $X$ and $Y$ are integer-valued.

In case anyone else is interested (a) is answered simply by seeing that $X+Y = V+W$ and therefore $E(V+W)=E(X+Y)=E(X)+E(Y)=E(V)+E(W)$. This simply uses linearity of expectation.

For (b), I believe that $P(X<Y) = P(n-X<n+1-Y)$ by symmetry, since both $X$ and $Y$ are binomially-distributed with $p=0.5$. So, my first question is: is this correct? I am not really sure how to justify this mathematically, but seems to be intuitively true because $P(X=k) = P(X=n-k)$ (although maybe this is wrong ;)).

For (c), the question suggests using part (b). So, I took that to mean $1 = P(X<Y) + P(Y>X) + P(X=Y)$. So, I took that to mean $$1 = P(X<Y) + P(Y>X) + P(X=Y) = P(X<Y) + P(n-X<n+1-Y) + P(X=Y) = 2P(X<Y) + P(X=Y).$$Therefore, $P(X<Y) = \frac{1-P(X=Y)}{2}$. I think then $P(X=Y) = \sum_{k=0}^{n} P(X=k)P(Y=k)$. I feel like I am now using that $X$ and $Y$ are integer-valued, but I am thinking there is a further simplification of this formula. Can someone help me get on the right track here? Just an additional note, is it possible to use part (a) at all in (b) and (c)? I don't really see how these are connected.

Thanks,

- J

Alice flips a fair coin n times and Bob flips another fair coin n + 1 times, resulting in independent $X \sim Bin(n, 1/2)$ and $Y \sim Bin(n+1, 1/2)$.

(a) Let $V = min(X,Y)$ and $W = max(X,Y)$. Find $E(V) + E(W)$. If $X=Y$ then $X=Y=V=W$.

(b) Show that $P(X<Y) = P(n-X<n+1-Y)$.

(c) Compute $P(X<Y)$. Hint: Use (b) and the fact that $X$ and $Y$ are integer-valued.

In case anyone else is interested (a) is answered simply by seeing that $X+Y = V+W$ and therefore $E(V+W)=E(X+Y)=E(X)+E(Y)=E(V)+E(W)$. This simply uses linearity of expectation.

For (b), I believe that $P(X<Y) = P(n-X<n+1-Y)$ by symmetry, since both $X$ and $Y$ are binomially-distributed with $p=0.5$. So, my first question is: is this correct? I am not really sure how to justify this mathematically, but seems to be intuitively true because $P(X=k) = P(X=n-k)$ (although maybe this is wrong ;)).

For (c), the question suggests using part (b). So, I took that to mean $1 = P(X<Y) + P(Y>X) + P(X=Y)$. So, I took that to mean $$1 = P(X<Y) + P(Y>X) + P(X=Y) = P(X<Y) + P(n-X<n+1-Y) + P(X=Y) = 2P(X<Y) + P(X=Y).$$Therefore, $P(X<Y) = \frac{1-P(X=Y)}{2}$. I think then $P(X=Y) = \sum_{k=0}^{n} P(X=k)P(Y=k)$. I feel like I am now using that $X$ and $Y$ are integer-valued, but I am thinking there is a further simplification of this formula. Can someone help me get on the right track here? Just an additional note, is it possible to use part (a) at all in (b) and (c)? I don't really see how these are connected.

Thanks,

- J

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