Matematik
Normal distribution
Hi!
In the past couple of hours I have solved a lot of problems regarding the normal distribution. I have a basic understanding of the distribution and I am doing just fine when using the probability function as well as the cumulative probability function. I am struggling though when I have to use the moment generation function (m.g.f) - so I am hoping you can help me!
Problem:
Given , find E(X3) (LaTeX does not do good with ~, sorry)
(a) using the m.g.f
Now the m.g.f is defined as
I should then be easy to find the correct moment by inserting. The problem is that we do not know the variable t - how do we find this variable? My book is very unclear about this matter.
I read about the standard normal distribution which is characterized by having and a m.g.f:
which we can expand by a taylor series expansion around 0 to find the even-order moments (all odd-order moments equals zero). But this does not help me with my problem.
Perhaps I have missed something totally basic or as you might say: "du kan ikke se skoven for bare træer". ;) Can anyone help me clear this up? I would be very thankful. :)
Svar #1
13. oktober 2010 af Henriksorensen (Slettet)
I spell bad..
Moment GENERATING function (m.g.f) OF COURSE.
Svar #2
13. oktober 2010 af Andersen11 (Slettet)
#1
Det er måske nemmere så at holde diskussionen på dansk.
Denne artikel på wiki http://en.wikipedia.org/wiki/Moment-generating_function kan måske være til hjælp.
Svar #3
13. oktober 2010 af Henriksorensen (Slettet)
I am afraid my danish is not good enough (yet) to discuss such hardcore matters. :) An explanation in english would probably be easier to understand. I hope it is not too much to ask. I figured that since all your academic books are english anyway, it would do no difference.
I read the article. Unfortunately, it did not help me solve the problem.
Svar #4
13. oktober 2010 af Andersen11 (Slettet)
#3
Well, your English sucks, too, so who cares?
The moment generating function is called MX(t) in the wiki article mentioned in #2. As you also mention, for a normal distribution N(μ,σ2) it is
MX(t) = etμ+(1/2)σ^2·t^2 (1)
It is also mentioned that
MX(t) = ∑k=0∝ mktk/(k!) ,
where mk = E(Xk) is the k'th moment .
Hence, by forming the Taylor expansion of expression (1), we can immediately read off the k'th moment from the power series. In particular, m3 = E(X3) can be calculated for μ=2 and σ2=1
Svar #5
13. oktober 2010 af Henriksorensen (Slettet)
Thanks! Your attitude was very helpful! Oh wait, your skills in statistics were, your attitude was just .. well, I guess you know.
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