Fair Inf
Fair Inf
power series question?
A fair coin is tossed repeatedly. The probability that the first head occurs on the nth toss is P(n)=(1/2)^n. When this action is repeated many times, the average # of tosses required until the first head is:
E(n)=sum from 1 to inf n*P(n)
This # is called the expected value of n
A) use the power series: 1/1-x = sum from 0 to inf of x^n, for |x|<1 to get a power series representation for f(x)=1/(1-x)^2 and give interval of convergence
B) what is the relation between E(n) and the series found in (a)?
C) what is the expected value of n? Be sure to give specific reason why you can make your conclusion.
I am completely clueless on this. If someone could just help me on each part. Someone has tried explaining it but I still do not get it. Thanks
A) You are given that if |x|<1, 1/(1-x) = Σ{n=0 to inf} x^n
There is a theorem that says if f(x) = Σ a[n] x^n where a[n] is a sequence of constants, then f'(x) = Σ d(a[n] x^n)/dx and the radius of convergence is the same as for f(x).
Notice that if f(x) = 1/(1-x) then f'(x) = 1/(1-x)^2. Therefore
1/(1-x)^2 = f'(x) = Σ{n=0 to inf} d(x^n)/dx
= d(1)/dx + Σ{n=1 to inf} d(x^n)/dx
= Σ{n=1 to inf} nx^(n-1)
B) E(N) = Σ{n=1 to inf} nP(n) = Σ{n=1 to inf} n (1/2)^n
= Σ{n=1 to inf} n (1/2) * (1/2)^(n-1)
= (1/2) Σ{n=1 to inf} n (1/2)^(n-1)
Notice that the summation looks exactly like the summation in part A when you set x=1/2.
So E(N) = 1/2 f'(1/2).
C) Since f'(1/2) = 1/(1-1/2)^2 = 4
E(N) = 2
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