Expected Value (Algebra)

Original post by JKITFC

Please see attached screenshot!

What happens between 'starting from x=1' to 'using y=x-1'
I know its times by (1-n) and divide by (1-n) - but I don't know why the 'to the power of x' goes to x-1 and how the x within the sum goes to (y+1)..

All help appreciated..

It's just using the rule of indices:

a^n = a^{n-1} \times a

.

In this case:

(1-\pi)^x = (1-\pi)^{x-1} \times (1-\pi)

.

Since

(1-\pi)

is just a constant, you can pull it out of the sum:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\sum x(1-\pi)^x = \sum x (1-\pi)^{x-1} (1-\pi) = (1-\pi)\sum x(1-\pi)^{x-1}\end{equation*}

Reply 2

OP

Original post by Zacken

It's just using the rule of indices:

a^n = a^{n-1} \times a

.

In this case:

(1-\pi)^x = (1-\pi)^{x-1} \times (1-\pi)

.

Since

(1-\pi)

is just a constant, you can pull it out of the sum:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\sum x(1-\pi)^x = \sum x (1-\pi)^{x-1} (1-\pi) = (1-\pi)\sum x(1-\pi)^{x-1}\end{equation*}

Perfect - thankyou

Reply 3

Original post by JKITFC

Perfect - thankyou

And for the next part, think about it just like a "integration substitution"

The sub

y=x-1

means that when

x=1

, we have

y = 1-1 = 0

(that's the bottom limit sorted) and the top limit is

y = \infty -1 = \infty

so that stays the same.

Then, the

y = x-1

means that the

(1-\pi)^{x-1} = (1-\pi)^y

and the

x = y + 1

by re-arranging the sub

y=x-1

by adding 1 to both sides of the equation.

Reply 4

OP

Screen Shot 2016-05-02 at 15.00.21.png106.6KB

Thanks - that part clicked just after I posted!

I get the fact that the second part of the addition is a PMF so equals one, and the fact the first part = E(X).
How does (Y+1) go to y(1-n) - is it becasue the (1-n) outside the sum multiplies with the (y+1) inside?

Reply 5

Original post by JKITFC

Thanks - that part clicked just after I posted!

I get the fact that the second part of the addition is a PMF so equals one, and the fact the first part = E(X).
How does (Y+1) go to y(1-n) - is it becasue the (1-n) outside the sum multiplies with the (y+1) inside?

If you're asking about

(1-\pi)[E(X) + 1]

, then yeah - it's just multipling out, just like

a(b+c) = ab + bc

; we have:

(1-\pi)[E(X) + 1] = (1-\pi)E(X) +(1-\pi)1

(edited 7 years ago)

Reply 6

OP

Original post by Zacken

If you're asking about

(1-\pi)[E(X) + 1]

, then yeah - it's just multipling out, just like

a(b+c) = ab + bc

; we have:

(1-\pi)[E(X) + 1] = (1-\pi)E(X) +(1-\pi)1

I mean from the very top line of the second screenshot.
How does (Y+1) go to y(1-n) - so how do they make the (y+1) at the bottom of the first screenshot into the E(X) in the top line of the second screenshot?

(edited 7 years ago)

Reply 7

Original post by JKITFC

I mean from the very top line of the second screenshot.

Oh, I think I understand what you mean.

Unparseable latex formula:

\displaystyle [br]\begin{equation*}(y+1)(1-\pi)^y \pi = (y)(1-\pi)^y \pi + (1)(1-\pi)^y \pi \end{equation*}

By expanding the

(y+1)(a) = ay + a

where

a = (1-\pi)^y \pi

.

So:

Unparseable latex formula:

\displaystyle[br]\begin{align*}(1-\pi)\sum(y+1)(1-\pi)^y y &= (1-\pi) \sum \big( (y)(1-\pi)^y \pi + (1)(1-\pi)^y \pi \big) \\ & = (1-\pi)\bigg[ \sum y(1-\pi)^y \pi + \sum (1-\pi)^y \pi \bigg]\end{equation}

Reply 8

OP