# Yes, it's another stats question

Discussion in 'Mathematics' started by beedge, Feb 25, 2014.

1. ### beedgeNew commenter

A biased coin has a probability p, that it gives a tail when it is tossed. The random variable T is the number of tosses up to and including the second tail.

(a) state the distribution of T.

NB (2, r)

(b) Show that p(T = t) = (t-1). (1-p)^t-2. p^2 for t > 2

(c) Hence show that 1/(T - 1) is an unbiased estimator of p.

It's part c - I really don't know what I supposed to do here.

2. ### jeremyb66

Well, the very first question is "Can you write down in words/symbols what it means for A to be an unbiased estimator of B" - in other words, do you know what you should be aiming for?

3. ### Come_to_TSR

In part (b) the formula given is valid for t>1.

This will be useful when you apply jeremyb66's hint.

4. ### beedgeNew commenter

Jeremyb66 - no this is the first problem. I think if I at least knew what I was aiming to do, then I would be able to make a good start.

I've been getting to grips with questions where you have to show that things like:

cX1 + (1-c)X2 forms an unbiased estimate of a population mean

or, if f(x) = 1 / k 1 < x < k+ 1, finding an unbiased estimator for k

But because I don't understand it really, as soon as there's a new situation (like this one), I get stuck.

I know that the expected value of T is 2/p. Am I trying to show that this, equals the same thing, but when you substitute 1/(T-1) into something?

As you see, I really don't have a clue.

Comtre_To_TSR - yes sorry about that. I was just too lazy to find this symbol: ?

5. ### jeremyb66

beedge, I was being deliberately vague because I was hoping that somewhere in your textbook / stats notes you would have 2 things:

a definition of "unbiased estimator" in terms of expectations

a definition of "expectation" in terms of values and probabilities

so I thought a little nudge in the right direction might help

Do these suggestions ring any bells?

Just out of interest, what do you use as a standard textbook for your course?

6. ### beedgeNew commenter

Ok, so I know the following:

An unbiased estimator of a population parameter has an expectation equal to the population parameter: if 'a' is a parameter of a population then the sample statistic 'A' is an unbiased estimator of a if E(A) = a.

This is according to the Cambridge Maths HL options book: stats and probability.

In my question, my unbiased estimator is 1 / (t - 1)

So am I trying to show the E ( 1 / (t - 1)) = p ?

If so, how ???

7. ### Come_to_TSR

In that same book I'm sure you will find the definition of E(X).

8. ### jeremyb66

As Come_to_TSR says, dig out your book's definition of E(X) for a random variable X. In your case you will find that it involves the expression that you worked out for p(T = t) earlier. Plug this expression into your definition and you should find that something magical happens

9. ### beedgeNew commenter

Are you 2 sitting next to each other, revelling in my misfortune??

Where's the maths teacher solidarity here??

Honestly...

"The expected value that you would expect to observe on average if you repeat the experiment an infinite number of times. The formula we use to determine the expected value can be simply understood with an example..."

Then they give a simple example given for a discrete random variable.

Followed by the formula: ?x P(x), which applies for all discrete distributions.

That's actually from the "core" part of the course, and not the "options" part.

So exactly how do I combine that with this:

p(T = t) = (t-1). (1-p)^t-2. p^2

and this:

p = 1/(T-1)

or that the expected value for our particular question is 2/p??

What am I trying to show = what?

10. ### Come_to_TSR

Not at all. The idea is to make you more independent.

Thinking about a simpler case that you're probably familiar with: E(X^2) = &sum;x^2 P(X=x)

Your problem is much the same.

From E(X)=&sum;x P(X=x) we can get E(1/(T-1)) = &sum; 1/(t-1) P(T=t) ( t &ge; 2 )

The sum is...nice.

11. ### jeremyb66

Sorry you feel our efforts are unsupportive. I genuinely thought that you would be able to complete the problem from the hints given. I have to confess a personal aversion to statistics/prob questions, but this is actually "my sort of probability question" because it's really all algebra in disguise!

Come_to_TSR has now given you the required formula for E(1/(T-1)). The factor 1/(t-1) in the summation cancels your original (t-1) factor so you are left with summing p*2.(1-p)^t-2.

You can take the p^2 outside the summation and the sum itself is the sum of (1-p)^t-2 from t= 2 to infinity. If you actually write out the first few terms you will find it easier to see what's going on - it's just a geometric series.

You are aiming to show that E(1/(T-1)) = p which is what I was hinting at originally.

Sorry again if I was being too obtuse.