Math problems!

[cryofrzd]

This part of Probabilities is actually on the schedule of my upcoming math exam, so I'll make a mental note that this equation exists ^^"

Good luck on the exam!

 

I'll take Mathematical statistics exam next month, for the third time, this time however I hope to get a passing grade.

[Mephisto]

This is a cry for help in my sleepy state as I realize that my skill in calculus is insufficient to properly perform in the statistics case I'm currently taking (it's the first fucking stats class I have to take too wtf?) 

 

どうか助けてください 

 

If anyone knows anything about stuff like means/variances of continuous distributions, pdfs, cdfs, whatever, please message me or something or post here. help needed ;_;

 

Basically I have no idea what I'm doing anymore. The easier questions are things like 

 

Let a and b be positive integers with a <= b, and let X be a random variable that takes as values with equal probability, the power of 2 in the interval [2^a, 2^b]. Find the expected value and the variance of X. 

 

I think I'm supposed to do something like integral of xf(x)dx from (2^(-infinity),2^a) with f(x) = 2^x or something? Fuck I don't even know what's going on anymore. I don't even know how to integrate x * 2^x anymore. 

[Down]

If anyone knows anything about stuff like means/variances of continuous distributions, pdfs, cdfs, whatever, please message me or something or post here. help needed ;_;

Can't give you more than wikipedia there. The definitions on wikipedia are usually pretty solid though. Once you got your definitions straight and you identify correctly what situation you have it shouldn't be too big of a deal.

 

Let a and b be positive integers with a <= b, and let X be a random variable that takes as values with equal probability, the power of 2 in the interval [2^a, 2^b]. Find the expected value and the variance of X. 

 

I think I'm supposed to do something like integral of xf(x)dx from (2^(-infinity),2^a) with f(x) = 2^x or something? Fuck I don't even know what's going on anymore. I don't even know how to integrate x * 2^x anymore.

Yeah, your head blew up here. Get some sleep

 

Your variable is taking discrete values, so all you have to deal with is finite sums and not integrals. Expected value would be sum from a to b of (1/(a-b )*2^n (sum of probabilities of values times values) which is trivial. Variance would similarly be sum from a to b of (1/(a-b )*n*2^n. Less trivial but pretty sure there's a way to make a formula out of this. Can't be bothered to do it but try out for the first few n then prove it by induction.

[Mephisto]

How did you get the probability of values?

 

if b = 3 a = 1

 

then your possible values are 2^1, 2^2, 2^3 

 

so the possibility of each would be 1/3 

 

1/(a-b ) = 1/(1-3) = 1/(-2) = -1/2 

 

if b = 3, a = -1 

possible values are 2^-1, 2^0, 2^1, 2^2, 2^3

possibility = 1/5 

1/(a- b ) = 1/(-1 - 3) = 1/(-4) = -1/4

 

maybe 1/(b-a + 1) ?

 

if b = 5 a = 3

2^3 2^4 2^5

1/(5-3+1) = 1/3

 

so

 

EX = sum from n=a to b of  2^n * (1/(b-a+1)) or something? 

[Down]

 

maybe 1/(b-a + 1) ?

I didn't count right and inverted the two, my bad. If a<=b then yes it's this.