1. ## Statistics

It's a geometric distribution but suppose we didn't know that

$\\\text{Find, via first principles, }Var(X)\\ \text{if }Pr(X=k)=\frac23 \left(\frac13\right)^{k-1}\, k =1,2,\dots$

$\text{Proven: }\mathbb{E}[X]=\frac32$

2. ## Re: Statistics

Originally Posted by leehuan
It's a geometric distribution but suppose we didn't know that

$\\\text{Find, via first principles, }Var(X)\\ \text{if }Pr(X=k)=\frac23 \left(\frac13\right)^{k-1}\, k =1,2,\dots$

$\text{Proven: }\mathbb{E}[X]=\frac32$
$\noindent One way is to essentially first derive the formula for \sum_{k\geq 0} k^{2}z^{k} (|z| < 1) by applying the operator z\frac{\mathrm{d}}{\mathrm{d}z} a couple of times to the geometric series \sum _{k\geq 0} z^{k} = \frac{1}{1-z} (|z| < 1).$

3. ## Re: Statistics

$\noindent The final answer will be \frac{q}{p^{2}}, where p is the success'' probability parameter, and q = 1-p. In your case, p = \frac{2}{3}.$

5. ## Re: Statistics

Originally Posted by Confound
Explain what? Every single detail? InteGrand's sum derivation?

6. ## Re: Statistics

Quick question regarding probability spaces:

$\text{Is the convention to take }\mathbb{P}(A) \ge 0\text{ or }\mathbb{P}(A) > 0?$

7. ## Re: Statistics

Originally Posted by leehuan
Quick question regarding probability spaces:

$\text{Is the convention to take }\mathbb{P}(A) \ge 0\text{ or }\mathbb{P}(A) > 0?$
Greater than or equal to. Events can have 0 probability.

8. ## Re: Statistics

leehuan - what would you do if InteGrand decides to take a 6 month sabbatical?

9. ## Re: Statistics

Originally Posted by Drongoski
leehuan - what would you do if InteGrand decides to take a 6 month sabbatical?
I don't know; InteGrand can do as he wants. But why are you asking me this?

10. ## Re: Statistics

Originally Posted by leehuan
I don't know; InteGrand can do as he wants. But why are you asking me this?
Just teasing you. And noting that InteGrand has been helping you so much.

11. ## Re: Statistics

Originally Posted by Drongoski
Just teasing you. And noting that InteGrand has been helping you so much.
Very true. So grateful.

12. ## Re: Statistics

I wasn't taught the hypergeometric distribution properly so can someone walk me through how to use it? Here's my question if it helps to refer to it.

A factory produces 80 items in a batch. To test if the batch is defective, an acceptance sampling scheme is adopted: a random sample of 10 items is selected, and if 2 or more items don’t meet customer specifications, the batch is considered defective.

If there are actually 11 defective items in the batch,

1i) What is the probability that 2 sampled items are defective?
ii) What is a general formula for x sampled items being defective?

13. ## Re: Statistics

Doing this for the first time for a very very long long time. Not sure if correct.

$P(X = k) = \frac{{\binom K k } {\binom {N-K} {n-k}}}{\binom N n} \\ \\Here N = 80, K = 11, n = 10 and for (i) k = 2 \\ \\ So: P(X = 2) = \frac {\binom {11} 2 \binom {80-11}{10-2} }{\binom {80} {10}} \\ \\ For k = x: P(X = x) = \frac {\binom {11} x \binom {80 - 11} {10-x}}{\binom {80} {10}}$

14. ## Re: Statistics

Originally Posted by Drongoski
Doing this for the first time for a very very long long time. Not sure if correct.

$P(X = k) = \frac{{\binom K k } {\binom {N-K} {n-k}}}{\binom N n} \\ \\Here N = 80, K = 11, n = 10 and for (i) k = 2 \\ \\ So: P(X = 2) = \frac {\binom {11} 2 \binom {80-11}{10-2} }{\binom {80} {10}} \\ \\ For k = x: P(X = x) = \frac {\binom {11} x \binom {80 - 11} {10-x}}{\binom {80} {10}}$
I think the expressions are right.

But I'm not sure what the parameters mean either. This question is pretty much an example but for an arbitrary scenario how would I be able to tell what the parameters (N, K, n) actually meant?

15. ## Re: Statistics

Originally Posted by leehuan
I think the expressions are right.

But I'm not sure what the parameters mean either. This question is pretty much an example but for an arbitrary scenario how would I be able to tell what the parameters (N, K, n) actually meant?
The parameters are as follows:

• N is the total population
• K is the number of "tagged" objects (defective objects in your example)
• n is the size of our sample.

The hypergeometric distribution pmf Drongoski wrote (in terms of the parameters N, K, n) then gives the probability that our sample has exactly k "tagged" (defective) objects present, under the assumption that we are sampling without replacement. This follows from basic combinatorics.

16. ## Re: Statistics

The reason for terms like "population" and "tagged" is that one place this distribution comes up is in ecology when we tag some members of an animal population (like a fish population) and then later draw (without replacement) a random sample from the animal population and count how many are tagged. This can be used to try and estimate the total population for example (it is sometimes known as the "capture-recapture method", and you can read more about it here: https://en.wikipedia.org/wiki/Mark_and_recapture).

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