Proof of the expected value of the geometric distribution.

This is a silly question, but I'm struggling to find anything online.

Does anyone have a copy of the OCR S1 textbook to hand? I want to check if it has a proof of the expected value of the geometric distribution in it.

I seem to remember an elegant proof based off the common ratio of a geometric sequence, however the only proofs I can find online that use that idea also use differential calculus. Long story short, the only place I can think of where I would've seen that proof would've been in the OCR S1 textbook, which I no longer own a copy of.

So, naturally, I'm here to ask if that's even the case, what proof does the book have?

Reply 1

TeeEm

19

Original post by BecauseFP

This is a silly question, but I'm struggling to find anything online.

Does anyone have a copy of the OCR S1 textbook to hand? I want to check if it has a proof of the expected value of the geometric distribution in it.

I seem to remember an elegant proof based off the common ratio of a geometric sequence, however the only proofs I can find online that use that idea also use differential calculus. Long story short, the only place I can think of where I would've seen that proof would've been in the OCR S1 textbook, which I no longer own a copy of.

So, naturally, I'm here to ask if that's even the case, what proof does the book have?

I will write one by tomorrow evening if you cannot find it or no one else provides it

Reply 2

DFranklin

18

Original post by BecauseFP

This is a silly question, but I'm struggling to find anything online.

Does anyone have a copy of the OCR S1 textbook to hand? I want to check if it has a proof of the expected value of the geometric distribution in it.

I seem to remember an elegant proof based off the common ratio of a geometric sequence, however the only proofs I can find online that use that idea also use differential calculus. Long story short, the only place I can think of where I would've seen that proof would've been in the OCR S1 textbook, which I no longer own a copy of.

So, naturally, I'm here to ask if that's even the case, what proof does the book have?

I think the following is most elegant

Let X be geometrically distributed with parameter p (and to remove anambiguity, X is the number of trials*until* successt, so X>0).

On the first trial, we have success with probability p (in which case we need no extra trials). Otherwise the number of additional trials before success is geometrically distributed with probability p.

Thus E[X] = 1 + (1-p)E[X], and so ,E[X] = 1/p.

Reply 3

TeeEm

19

Original post by BecauseFP

This is a silly question, but I'm struggling to find anything online.

Does anyone have a copy of the OCR S1 textbook to hand? I want to check if it has a proof of the expected value of the geometric distribution in it.

I seem to remember an elegant proof based off the common ratio of a geometric sequence, however the only proofs I can find online that use that idea also use differential calculus. Long story short, the only place I can think of where I would've seen that proof would've been in the OCR S1 textbook, which I no longer own a copy of.

So, naturally, I'm here to ask if that's even the case, what proof does the book have?

If DFranklin's proof suffices for your course is all good, if not I can post an elementary one

Reply 4

BecauseFP

OP

2

Original post by TeeEm

If DFranklin's proof suffices for your course is all good, if not I can post an elementary one

The elementary one sounds interesting,

Reply 5

TeeEm

19

Original post by BecauseFP

The elementary one sounds interesting,

here you go

Original post by BecauseFP

The elementary one sounds interesting,

Another alternative:
Let

X \sim Geo(p)

Then:

E[X]=\displaystyle\sum_{r=1}^{\infty} rP(X=r)

But we know that

P(X=r)=(1-p)^{r-1}p

.

\Rightarrow E[X]=\displaystyle\sum_{r=1}^{\infty} r(1-p)^{r-1}p=p\displaystyle\sum_{r=1}^{ \infty} r(1-p)^{r-1}[br]

Let

q=1-p

and sub in:

E[X]=p\displaystyle\sum_{r=1}^{\infty} rq^{r-1}

Notice that:

\displaystyle\sum_{r=1}^{\infty} rq^{r-1}=\dfrac{d}{dq} \left( \displaystyle\sum_{r=1}^{\infty} q^r \right)

.
Aside:

Spoiler

Hence we have:

E[X]=p\dfrac{d}{dq} \left( \displaystyle\sum_{r=1}^{\infty} q^r \right)

Using the formula for the sum of a geometric series (which applies as

|q|<1

.

E[X]=p\dfrac{d}{dq} \left( \dfrac{1}{1-q} \right) =p\dfrac{1}{(1-q)^2}

Recalling that

p=1-q \Rightarrow q=1-p

gives us that:

E[X]=p\dfrac{1}{p^2}=\dfrac{1}{p}. \ \ \ \square

(edited 8 years ago)

Reply 7

DFranklin

18

Nothing wrong with either of these proofs, but I fail to see why they are more elementary than one that's literally 1 line of calculation (and a little more explanation).

Reply 8

TeeEm

19

Original post by DFranklin

Nothing wrong with either of these proofs, but I fail to see why they are more elementary than one that's literally 1 line of calculation (and a little more explanation).

In my case I wrote the proof as I needed one to add to my own resources.
(Now I have two more proofs to add.)
The OP will decide what is complicated and what is elementary.
Besides offering many alternatives educates many others watching this thread.

Reply 9

HeavisideDelts

3

Original post by DFranklin

Nothing wrong with either of these proofs, but I fail to see why they are more elementary than one that's literally 1 line of calculation (and a little more explanation).

What she said:

Original post by TeeEm

In my case I wrote the proof as I needed one to add to my own resources.
(Now I have two more proofs to add.)
The OP will decide what is complicated and what is elementary.
Besides offering many alternatives educates many others watching this thread.

Reply 10

TeeEm

19

Original post by HeavisideDelts

What she said:

she?
I thought I was a he ever since I was a little boy....

Reply 11

HeavisideDelts

3

Original post by TeeEm

she?
I thought I was a he ever since I was a little boy....

don't lie to us caitlyn

Reply 12

Zacken

22

To the two posters above, I don't think DFranklin is saying that the alternatives are wrong or useless, just that he can't see why they are more 'elementary' than his, which I'm agreeing with, the one line proof is very elegant.