Probability proof question?

Question

Consider a random variable with a geometric distribution p(y) = q^y-1 * p (y= 1, 2, 3, .....)

a) Show that Y has a distribution function F(y) such that F(i) = 1- q^i, i = 0, 1, 2, 3.....

In general, F(y) = 0 for y<0, or 1-q^i for i<= y <=i+1

b) Show that the preceding function has the properties:

1)F(- infinity) = lim (y->-infinity) F(y) = 0

2) F(infinity) = lim (y->infinity) F(y) = 1

F(y) is a nondecreasing fucntion.

I don't see what I need to do for a)

For b), they all seem obvious, but I just don't see how to write it formally.

kb · Accepted Answer

a) Since p(y) is discrete, for y = 1, 2, ... :

F(y) = Σ(k = 1 to y) p(k)

........= Σ(k = 1 to y) q^(k-1) * p

........= p * Σ(k = 1 to y) q^(k-1)

........= p (1 - q^y)/(1 - q)

........= p (1 - q^y)/p, since p + q = 1

........= 1 - q^y.

The other portions for F should be clear from this point.

b) Assuming that q is in (0, 1), we have lim(n→ ±∞) q^n = 0.

So, lim(y→ -∞) F(y) = lim(y→ -∞) 0 = 0,

and lim(y→ -∞) F(y) = lim(y→ -∞) (1 - q^y) = 1 - 0 = 0.

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Let x > y be real numbers.

If x, y < 0, then F(x) - F(y) = 0 - 0 = 0.

If x ≥ 0 and y < 0, then F(x) - F(y) = F(x) - 0 > 0.

If x, y ≥ 0, then F(x) = p^m - 1 and p^n - 1 for some integers m in [x, x+1) and n in [y, y+1).

Note that m ≥ n.

==> F(x) - F(y) = (1 - q^m) - (1 - q^n) = q^n - q^m = q^n (1 - q^m) ≥ 0, since q is in (0, 1).

In any case, x > y ==> P(x) ≥ P(y).

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I hope this helps!

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