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).

17. ## Re: Statistics

Any tips for catching up on stats... like any resources you use?

Thanks I am behind lol.

18. ## Re: Statistics

Can't deny it. I'm also a fair bit behind.

Been cramming a lot of the course pack tbh.

19. ## Re: Statistics

Originally Posted by leehuan
Can't deny it. I'm also a fair bit behind.

Been cramming a lot of the course pack tbh.
what course pack? the questions that are on moodle or?

20. ## Re: Statistics

Originally Posted by Flop21
what course pack? the questions that are on moodle or?
Course pack is just what I use to call the online 'textbook' thing. It's like what we had in first year except it's just not printed.

21. ## Re: Statistics

is there an easy way to find the probability mass function??

right now im just looking at example P(X=x) like P(X=1... etc.) and trying to find a pattern

but this is hard for tricky ones

22. ## Re: Statistics

Originally Posted by Flop21
is there an easy way to find the probability mass function??

right now im just looking at example P(X=x) like P(X=1... etc.) and trying to find a pattern

but this is hard for tricky ones
Got an example question?

23. ## Re: Statistics

Originally Posted by He-Mann
Got an example question?
A box contains four red and two black balls. Two balls are drawn. Let X be
the number of red balls obtained. Find fX(x)

Okay so if I haven't screwed up, P(X=0) = 1/15, P(X=1) = 4/5, P(X=2) = 2/5, and any other value of X, P = 0.

So how do I put this into a proper answer for this question?

24. ## Re: Statistics

Originally Posted by Flop21
A box contains four red and two black balls. Two balls are drawn. Let X be
the number of red balls obtained. Find fX(x)

Okay so if I haven't screwed up, P(X=0) = 1/15, P(X=1) = 4/5, P(X=2) = 2/5, and any other value of X, P = 0.

So how do I put this into a proper answer for this question?
$\noindent Those answers aren't right (assuming sampling without replacement). Note your pmf values don't sum to 1, which they should. This'll just be a hypergeometric distribution, but you don't really need to know that here. We have, letting f(k) \equiv f_{X}(k) = \mathbb{P}\left(X = k\right),$

$\mathbb{P}\left(X = 0\right) = \frac{1}{\binom{6}{2}} = \frac{1}{15} \quad (\text{so you got this one right})$

$\mathbb{P}\left(X = 1\right) = \frac{2\times 4}{\binom{6}{2}} = \frac{8}{15}$

$\noindent and$

$\mathbb{P}\left(X = 2\right) =\frac{\binom{4}{2}}{\binom{6}{2}} = \frac{6}{15} = \frac{2}{5} \quad (\text{you got this one right too}).$

$\noindent (Perhaps you just typoed one of your answers since you got the other two right.) So the pmf is f(0) = \frac{1}{15}, f(1) = \frac{8}{15} and f(2) = \frac{2}{5}, and of course f(k) = 0 for k \notin \left\{0,1,2\right\}.$

25. ## Re: Statistics

Originally Posted by InteGrand
$\noindent Those answers aren't right (assuming sampling without replacement). Note your pmf values don't sum to 1, which they should. This'll just be a hypergeometric distribution, but you don't really need to know that here. We have, letting f(k) \equiv f_{X}(k) = \mathbb{P}\left(X = k\right),$

$\mathbb{P}\left(X = 0\right) = \frac{1}{\binom{6}{2}} = \frac{1}{15} \quad (\text{so you got this one right})$

$\mathbb{P}\left(X = 1\right) = \frac{2\times 4}{\binom{6}{2}} = \frac{8}{15}$

$\noindent and$

$\mathbb{P}\left(X = 2\right) =\frac{\binom{4}{2}}{\binom{6}{2}} = \frac{6}{15} = \frac{2}{5} \quad (\text{you got this one right too}).$

$\noindent (Perhaps you just typoed one of your answers since you got the other two right.) So the pmf is f(0) = \frac{1}{15}, f(1) = \frac{8}{15} and f(2) = \frac{2}{5}, and of course f(k) = 0 for k \notin \left\{0,1,2\right\}.$
yeah whoops typo'd something in my calculations so that stuffed up,

so your final sentence, is that enough for an answer or do I need to know how to put it in a form with x ?

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