Integrating normal from 0 to Inf and finding the third moment

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SUMMARY

The discussion centers on the challenge of integrating the third moment of a normal distribution, specifically the integral ∫ x^3 f(x) dx from 0 to ∞, where f(x) is the probability density function (pdf) of the normal distribution with mean (-σ^2/b) and variance σ^2. Participants suggest using integration by parts and substitutions like x^2 = u, but conclude that there is no analytic solution due to the resulting error function. The consensus is that a "nice" antiderivative does not exist for this integral, necessitating numerical methods or tables for area calculations.

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  • Understanding of normal distribution and its properties
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  • Basic concepts of probability density functions (pdf)
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Hejdun
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This is (perhaps) a tricky question regarding the moment of normal distribution.
Let f(x) be the pdf of normal distribution with mean (-σ^2/b) and variance σ^2, where b is just a constant. The goal is to solve the integral

∫ x^3 f(x) dx

integrating from 0 to ∞.

I am stuck. Any suggestions?
 
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Use integration by parts to express the problem as one in lower powers of X. The expressions involving the lower powers will have something to do with lower order moments.

If it is a normal distribution, aren't you integrating from -\infty to \infty?
 
Stephen Tashi said:
Use integration by parts to express the problem as one in lower powers of X. The expressions involving the lower powers will have something to do with lower order moments.

If it is a normal distribution, aren't you integrating from -\infty to \infty?


I will try integration by parts, but I am not sure if that would solve it.

No, unfortunately it is from 0 to ∞. Otherwise it would have been a standard result.
 
Since the exponent will contain multiples of x^2 and x, h'about trying a substitution like:

x^2=u , then x^3=u^(3/2) , etc.
 
It seems that there is no analyc answer to this integral since integration will result in an error function. But thanks anyway!
 
But the Error function is analytic! It is an entire function.
 
Herick said:
But the Error function is analytic! It is an entire function.

I think by analytic Hedjun means that the function has an antiderivative in a

"nice closed form" , not in the complex-analytic sense.

Notice that in this case f is assumed real-valued.
 
Last edited:
Well, is it e^x a closed form? If by closed form he means a convergent infinite series, then Err(x) is also 'closed form'. Right?
 
Herick said:
Well, is it e^x a closed form? If by closed form he means a convergent infinite series, then Err(x) is also 'closed form'. Right?

I assume s/he , means a 'nice' antiderivative ( a combination of sums, products of

known/common functions ) F that would allow you to calculate the area between any two

x,y , without having to use numerical methods , or tables. AFAIK there isn't any one.

Do you know of one? I assumed, since tables are used for the normal density, that

there isn't any "nice-enough" antiderivative. But maybe the OP can tell us what s/he

was looking for.
 
  • #10
Bacle2 said:
I assume s/he , means a 'nice' antiderivative ( a combination of sums, products of

known/common functions ) F that would allow you to calculate the area between any two

x,y , without having to use numerical methods , or tables. AFAIK there isn't any one.

Do you know of one? I assumed, since tables are used for the normal density, that

there isn't any "nice-enough" antiderivative. But maybe the OP can tell us what s/he

was looking for.

Yes, it is this that I was looking for, which I believe does not exist for the integral that started this thread. Thanks for your replies.
 

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