Suppose that fx xc 0 x 1 2 n otherwise Find c so that fx

Suppose that f(x) = x/c 0 x = 1, 2, ..., n otherwise. Find c so that f(x) represents a pdf, Find an expression for The CDF, F(x). Find E(X).

The pdf must sum up to 1.

Thus,

1/c + 2/c + 3/c ... + n/c = 1

which is an arithmetic series of common difference d = 1/c, first term a1 = 1/c, last term an = n/c, with n terms.

Thus,

Sum = (1/c + n/c) n / 2

Sum = (1 + n) n / (2c) = 1

Thus,

c = (1+ n) n / 2 [ANSWER]

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Summing the first x terms,

Sum = (1/c + x/c) x / 2

Sum = (1 + x) x / (2c)

Plugging in the c we got,

Sum = F(x) = (1 + x) x / [(1+ n) n] [ANSWER]

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E(x) = Sum (x f(x)) = Sum(x^2/c) = (1/c) Sum(x^2)

The sum of the first n squares is n(n+1)(2n+1)/6. Thus,

E(x) = (1/c)(n(n+1)(2n+1)/6)

E(x) = {2/[(1+ n) n]} (n(n+1)(2n+1)/6)

E(x) = {1/[(1+ n) n]} (n(n+1)(2n+1)/3)

E(x) = [n(n+1)(2n+1)]/[3(1+ n) n]

E(x) = (2n+1)/3 [ANSWER]