Let Xn be the random variable that counts the difference in the number
of tails and the number of heads when n coins are flipped. Assume that
the coins fair, that is, heads and tails occur with the same
probability equal to 1/2. Compute E[Xn] and Var[Xn].

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Request for Question Clarification by rbnn-ga on 06 Nov 2002 20:26 PST

What is "difference" here? If you get 5 tails and 8 heads, is the
difference 3 or -3?

4 is not enough for me to tackle this one though.

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Clarification of Question by math01-ga on 07 Nov 2002 07:50 PST

The
difference 3

Hint:

Assuming that a simple difference, which might be positive, zero, or
negative, is meant (see rbnn-ga's request for clarification), we know:

(# of heads) = n - (# of tails)

X_n = (# of tails) - (# of heads) = 2*(# of tails) - n

But E(# of tails) is n/2, so E(X_n) = ???

Similarly one can relate the variance of X_n to that of the simple
binomial distribution for # of tails.

regards, mathtalk-ga

Let k be the difference of tails and heads in n trials
(not not every k between 0 and n can be outcomes)
since, the prob of k and -k are the same for each possible difference k >0.
e.g when n = 5, possible difference are 5, 3, 1 ,-1,-3,-5
when n = 6,possible difference are 6, 4, 2, 0 ,-2,-4,-6)
 
Hence E(Xn) = 0 ( in fact, it is equal to (k-k)* Pr(Xn = k) 
                  with summation over possible k from 0 to n)

Use the formula Var(Xn) = E(Xn^2) - E(Xn)^2 = E(Xn^2)- 0 = E(Xn^2)
To find E(Xn^2) = S k^2 * Pr(Xn^2 = k^2)  (S means summation over possible k)
 when n is odd, S k^2 * Pr(Xn^2 = k^2) = (1/2)^n S k^2
                = (1/2)^n (1+3^2+5^2+..+n^2)
let n=2m+1, 1+3^2+5^2+..+n^2=(1+2^2+3^2+4^2+..+n^2)-(2^2+3^2+4^2+..+(2m)^2))
  (all summation from 1 to n minus even terms)
            = n(n+1)(2n+1)/6 - 4m(m+1)(2m+1)/6
            = [n(n+1)(2n+1)- 4m(m+1)n] /6
            = n [(n+1)(2n+1) -4m(m+1)] / 6
            = 2 n(m+1)[(2n+1)- 2m] / 6
            = n(m+1)(2m+3)/3  
            = (m+1)(2m+1)(2m+3)/ 3
 so E(Xn^2) = (1/2)^n *(m+1)(2m+1)(2m+3)/ 3 , when n = 2m+1

 when n is even, S k^2 * Pr(Xn^2 = k^2) = (1/2)^n S k^2
                = (1/2)^n (2^2+4^2+6^2+..+n^2)
                = (1/2)^n * 2^2 (1+ 2^2+ 3^2+..+(n/2)^2)
                
 Let n = 2m, use 1+ 2^2+ 3^2+..+ m^2 = m(m+1)(2m+1)/ 6
            we have   = (1/2)^(n-1)* m(m+1)(2m+1)/3
  

khwang

Hi, Kenny:

If math01's interpretation is for |X_n| instead (see rbnn's request
for clarification and math01's reply), notice that:

E(|X_n|^2) = E((X_n)^2)

So that part of your calculation would carry over.

It needs only the ingredient E(|X_n|) to complete the analysis.

regards, mathtalk-ga

To answer mathtalk:
 Thanks for your reminding me. Actually, to
 compute E(|Xn|) should be easier than doing E(|Xn|^2)
 
To find E(|Xn|) = S k * Pr( |Xn| = k) (S means summation over possible k) 
 when n is odd, S k * Pr(|Xn| = k) = (1/2)^n S k 
                = (1/2)^n (1+3+5+..+n) (as arthmetic series)
                = (1/2)^n [(n+1)/2]^2
                = (n+1)^2/(2^(n+2))
 when n is even S k * Pr(|Xn| = k) = (1/2)^n S k 
                = (1/2)^n (0+2+4+6+..+n) (as arthmetic series)
                = (1/2)^n * 2[1+2+3+..+n/2]
                = (1/2)^n* n/2(n/2+1)
                = (1/2)^(n+1) * n(n/2+1) (after simplify)

 Combine with E(Xn^2)as I did, we will get Var(Xn)

khwang