Varaiance of Normal Distribution

INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]= x^2*-(e^(-x^2/2))/x|(inf -inf) - INT(inf - inf) (2*x)*-(e^(-x^2/2))/x= x*-(e^(-x^2/2))|(inf -inf) + INT|(inf -inf) 2*e^(-x^2/2)dx= -x*e^(-x^2/2)|(inf -inf) + INT|(inf -inf) 2*e^(-x^2/2)dx= -x*e
  • #1
Bazman
21
0
When solving for the variance of the normal distribution one needs to evaluate the following integral:

INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

I proceed using integration by parts:

[-x.e^(-x^2/2)|(infin to -infin) + INT(-infin to infin) 2*e(-x^2/2)dx]

However apparently correct answer is:

[-x.e^(-x^2/2)|(infin to -infin) + INT(-infin to infin) e(-x^2/2)dx]

but I don't see how the 2 cancels?
 
Last edited:
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  • #2
The integral of xe^(-x^2)/2) is xe^(-x^2)/2), while the derivative of x is 1. I don't see where you got the 2 from.
 
  • #3
Hi, thanks:

First I'll take you through what I did before I follow your solution through fully:

When I tried to solve the following integral:

INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

To solve using integration by parts: INT uv` = uv - INT u'v
I take u=x^2
u`=2*x

v`=e^(-x^2/2)
v =-(e^(-x^2/2))/x

hence

INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

= x^2*-(e^(-x^2/2))/x|(inf -inf) - INT(inf - inf) (2*x)*-(e^(-x^2/2))/x

= x*-(e^(-x^2/2))|(inf -inf) + INT|(inf -inf) (2)*(e^(-x^2/2))

clearly first term equates to 0

INT|(inf -inf) (2)*(e^(-x^2/2))
=2*INT|(inf -inf) (e^(-x^2/2))

I know that INT|(inf -inf) (e^(-x^2/2)) = SQRT(2pi).
so I get final answer of 2*SQRT(2pi) but the correct answer is SQRT(2pi)






It seems from the above you wish to take:

u=x
u`=1

v`=x.e^(-x^2/2)
v =-e^(-x^2/2)

correct?

In which case:

INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

= x*-(e^(-x^2/2))|(inf -inf) - INT(inf - inf) (1)*-(e^(-x^2/2))

= x*-(e^(-x^2/2))|(inf -inf) + INT|(inf -inf) (1)*(e^(-x^2/2))

Which seems to give the correct answer of SQRT(2pi)

However I cannot find any error in my working above if you follow it through the answer still seems to be 2*SQRT(2pi)?
 
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  • #5
Hi Quasar,

Thanks for that it confirms that using u=x and dv/dx= x*exp(-x^2/2)dx in the integration by parts it gives the correct answer. Which I agree with.

However my remaining question is that how come when I work through using u=x^2 and dv/dx= exp(-x^2/2)dx in the integration by parts the calculation out by a factor of 2?

I just would like to know in case I face similar problems in future. To me it seems more "natural" to choose u=x^2 and I guess it must yield the same answer if worked through correctly. My question is where am I going worng at present?

I take u=x^2
u`=2*x

v`=e^(-x^2/2)
v =-(e^(-x^2/2))/x

hence

INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

= x^2*-(e^(-x^2/2))/x|(inf -inf) - INT(inf - inf) (2*x)*-(e^(-x^2/2))/x

= x*-(e^(-x^2/2))|(inf -inf) + INT|(inf -inf) (2)*(e^(-x^2/2))

clearly first term equates to 0

INT|(inf -inf) (2)*(e^(-x^2/2))
=2*INT|(inf -inf) (e^(-x^2/2))

I know that INT|(inf -inf) (e^(-x^2/2)) = SQRT(2pi).
so I get final answer of 2*SQRT(2pi) but the correct answer is SQRT(2pi)
 
  • #6
Bazman said:
Hi Quasar,

Thanks for that it confirms that using u=x and dv/dx= x*exp(-x^2/2)dx in the integration by parts it gives the correct answer. Which I agree with.

However my remaining question is that how come when I work through using u=x^2 and dv/dx= exp(-x^2/2)dx in the integration by parts the calculation out by a factor of 2?

I just would like to know in case I face similar problems in future. To me it seems more "natural" to choose u=x^2 and I guess it must yield the same answer if worked through correctly. My question is where am I going worng at present?

I take u=x^2
u`=2*x

v`=e^(-x^2/2)
v =-(e^(-x^2/2))/x
This is incorrect as you can see by differentiating your v. [tex]e^{-x^2/2}[/itex] does not have an elementary anti-derivative.

hence

INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]

= x^2*-(e^(-x^2/2))/x|(inf -inf) - INT(inf - inf) (2*x)*-(e^(-x^2/2))/x

= x*-(e^(-x^2/2))|(inf -inf) + INT|(inf -inf) (2)*(e^(-x^2/2))

clearly first term equates to 0

INT|(inf -inf) (2)*(e^(-x^2/2))
=2*INT|(inf -inf) (e^(-x^2/2))

I know that INT|(inf -inf) (e^(-x^2/2)) = SQRT(2pi).
so I get final answer of 2*SQRT(2pi) but the correct answer is SQRT(2pi)
 
  • #7
v`=e^(-x^2/2)
v =-(e^(-x^2/2))/x
wrong (take the derivative of your v and you won't get your v')!
you should use
v'=xe^(-x^2/2)
v=-e^(-x^2/2)
 
Last edited:

Related to Varaiance of Normal Distribution

What is the variance of a normal distribution?

The variance of a normal distribution is a measure of how spread out the data is from the mean. It tells us how much the data points deviate from the average value.

How is the variance calculated for a normal distribution?

The variance for a normal distribution is calculated by taking the average of the squared differences of each data point from the mean. This is represented by the formula σ² = Σ(x-μ)² / N, where σ² is the variance, x is each data point, μ is the mean, and N is the total number of data points.

What is the relationship between variance and standard deviation for a normal distribution?

The standard deviation is the square root of the variance, so they are closely related. Standard deviation tells us the average amount that the data points deviate from the mean, while variance gives us a more precise measure of this deviation.

Why is the variance of a normal distribution important?

The variance of a normal distribution is important because it helps us understand the distribution of data and make predictions about future data. It is also used in many statistical tests to determine the significance of results.

How does the variance change with different normal distributions?

The variance can vary greatly between different normal distributions. It is affected by the shape and spread of the data, with wider and more spread out distributions having a higher variance, while narrower and more concentrated distributions have a lower variance.

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