Three proofs.

**approx** · September 22nd 2008, 07:09 AM

Hi! I've got a few problems that I wonder if someone could help me with.

1) If Y is a discrete random variable that assigns positive probabilites to only the positive integers, show that.

I don't have a clue about how to solve this one.

2) Let Y denote a geometric random variable with probability of success p.
a) Show that for a positive integer a,

b) Show that for positive integers a and b

I've been trying a lot with this one, but don't really know how to prove this.

3) Find E[Y(Y-1)] for a geometric random variable Y by finding d^2/dq^2 for

I don't know about this one either. My suggestion is that the second derivative is equal to this:

, but how do I use that to find E[Y(Y-1)]?

Would be nice if someone could help me with any of these problems.

**awkward** · September 22nd 2008, 06:04 PM

Originally Posted by approx

Hi! I've got a few problems that I wonder if someone could help me with.

1) If Y is a discrete random variable that assigns positive probabilites to only the positive integers, show that.

I don't have a clue about how to solve this one.
[snip]

$E(Y)$
$= \sum_{j=1}^\infty j \,P(Y=j)$
$= \sum_{j=1}^\infty \sum_{k=1}^j P(Y=j)$
$= \sum_{k=1}^\infty \sum_{j=k}^\infty P(Y=j)$
$= \sum_{k=1}^\infty P(Y \geq k)$

**approx** · September 23rd 2008, 05:54 AM

Thanks for your help. But could you please motivate the steps? And how come you end up with the index starting with k=1? Is it just a generalised thing?

**awkward** · September 23rd 2008, 04:53 PM

Originally Posted by approx

Thanks for your help. But could you please motivate the steps? And how come you end up with the index starting with k=1? Is it just a generalised thing?

The trick is simple, once you see it: express the sum as a double sum, then interchange the order of summation.

To see how to get from
$\sum_{j=1}^\infty \sum_{k=1}^j$
to
$\sum_{k=1}^\infty \sum_{j=k}^\infty$
take a piece of graph paper, label the axes j and k, and blacken the (j,k) points involved in the first sum. Then think about summing things in the other direction -- first k and then j.

**mr fantastic** · September 24th 2008, 01:34 AM

Originally Posted by approx

Hi! I've got a few problems that I wonder if someone could help me with.

[snip]

2) Let Y denote a geometric random variable with probability of success p.
a) Show that for a positive integer a,

[snip]

You should know that $\Pr(Y = n) = (1 - p)^{n-1} p = q^{n-1} (1 - q)$ .

Then:

$\Pr(Y > a) = 1 - \Pr(Y \leq a)$ $= 1 - \left[ ~ \Pr(Y = 1) + \Pr(Y = 2) + \Pr(Y = 3) + \, ... \, + \Pr(Y = a) ~ \right]$

$= 1 - \left[ ~ (1 - q) + q ( 1- q) + q^2 (1 - q) \, ... \, + q^{a-1} (1 - q) ~ \right]$

$= 1 - (1 - q) \left[ 1 + q + q^2 \, ... \, + q^{a-1} \right]$

Sum the stuff in the square brackets using the formula for the sum of a geometric series. Simplify.

$= 1 - [1 - q^a] = q^a$ .

**mr fantastic** · September 24th 2008, 03:42 AM

Originally Posted by approx

]snip]

2) Let Y denote a geometric random variable with probability of success p.

[snip]

b) Show that for positive integers a and b

I've been trying a lot with this one, but don't really know how to prove this.

[snip]

$\Pr( Y > a + b ~ | ~ Y > a) = \frac{\Pr(Y > a + b ~ \text{and} ~ Y > a)}{\Pr(Y > a)} = \frac{\Pr(Y > a + b)}{\Pr(Y > a)}$

Use the result from a):

$= \frac{q^{a+b}}{q^a} = q^b$ .

**mr fantastic** · September 24th 2008, 03:56 AM

Originally Posted by approx

[snip]
3) Find E[Y(Y-1)] for a geometric random variable Y by finding d^2/dq^2 for

I don't know about this one either. My suggestion is that the second derivative is equal to this:

, but how do I use that to find E[Y(Y-1)]?

Would be nice if someone could help me with any of these problems.

By definition:

$E[Y(Y-1)] = \sum_{y=1}^{\infty} y (y-1) q^{y-1} p = q p \sum_{y=1}^{\infty} y(y-1) q^{y-2} = q (1 - q) \sum_{y=1}^{\infty} y(y-1) q^{y-2}$ .... (1)

Sum of geometric series: $S = \sum_{y=1}^{\infty} q^{y} = \frac{q}{1-q}$ .

Therefore: $\frac{d^2 S}{d q^2} = \sum_{y=1}^{\infty} y (y - 1) q^{y-2} = \frac{2}{(1 - q)^3}$ .... (2)

Substitute equation (2) into equation (1) and simplify.

You can check this result by noting that:

$E(Y(Y-1)] = E[Y^2] - E[Y]$ $= E[Y^2] - \left( E[Y] \right)^2 + \left( E[Y] \right)^2 - E[Y] = Var(Y) + \left( E[Y] \right)^2 - E[Y]$ .

And for a geometric distribution you should know that $E[Y] = \frac{1}{p} = \frac{1}{1 - q}$ and $Var[Y] = \frac{1-p}{p^2} = \frac{q}{(1 - q)^2}$ .

**approx** · September 25th 2008, 11:43 AM

Thanks both of you, Mr F and awkward, for your help! Much appreciated.

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