Log Normal Distribution wtf?

[HP]

I was looking it up on Wiki, which says it was this:

http://img464.imageshack.us/img464/1786/371418813ae6c9e52069fcbdt5.png

I've never seen the lognormal distribution before, but this can't be right can it? Is the x in the divisor a mistake? wtf?

linky (http://en.wikipedia.org/wiki/Log-normal_distribution)

[HP]

anyone?

the long normal distribution should not be undefined @ x=0

however the above equation is, right?

[HP]

okay I just checked this (http://mathworld.wolfram.com/LogNormalDistribution.html) website and it says the same. I will be in deep thought trying to understand why the function is 0 @ x = 0

any help appreciated..

edit: nevermind. exp(-infinity) is 0, not 1

all is well with the universe

[PairTheBoard]

The pdf for the Log Normal is non-zero only for x>0.

You should be able to see it's correct by looking at the change of variable y=ln(x), dy=dx/x, x=e^y and integrating your f(x) from 0 to X for its Cumulative Distribution. After the change of variable the integral should equate to the Integral from -infinity to ln(X) of p(y) where p(y) is the standard normal pdf.

PairTheBoard

[jay_shark]

This is a typo !! Ln x should be replaced with x and the denominator should not have an x .
This is the correct formula .

f(x) = 1/[sqrt(2*pi)sigma]*e^[-(x-mu)^2/(2*sigma^2)]

where sigma =standard deviation
mu is the mean and x takes on the values from negative infinity to positive infinity . Think of f(x) as the y-value and naturally x is on the x-axis . We can evaluate the height of this function by plugging in the appropriate numbers .

In this case x takes on the mean mu , then all you have to do is compute the standard deviation which is simple to do .

It is almost the exact same formula . The only difference is that they have lnx instead of x and that there is an x in the denominator which mine doesn't have .

[PairTheBoard]

Also, see this Link.

Log Normal (http://mathworld.wolfram.com/LogNormalDistribution.html)

PairTheBoard

[HP]

[ QUOTE ]
The pdf for the Log Normal is non-zero only for x>0.

[/ QUOTE ]

yeah yeah, I meant the limit as x--> 0 you nit /images/graemlins/grin.gif edit: uh, nevermind thought you said something else

jay shark: I think that's the normal distribution you are talking about there

[thylacine]

[ QUOTE ]
The pdf for the Log Normal is non-zero only for x>0.

You should be able to see it's correct by looking at the change of variable y=ln(x), dy=dx/x, x=e^y and integrating your f(x) from 0 to X for its Cumulative Distribution. After the change of variable the integral should equate to the Integral from -infinity to ln(X) of p(y) where p(y) is the standard normal pdf.

PairTheBoard

[/ QUOTE ]

I 2nd this.

[PairTheBoard]

[ QUOTE ]
This is a typo !! Ln x should be replaced with x and the denominator should not have an x .
This is the correct formula .

f(x) = 1/[sqrt(2*pi)sigma]*e^[-(x-mu)^2/(2*sigma^2)]

where sigma =standard deviation
mu is the mean and x takes on the values from negative infinity to positive infinity . Think of f(x) as the y-value and naturally x is on the x-axis . We can evaluate the height of this function by plugging in the appropriate numbers .

In this case x takes on the mean mu , then all you have to do is compute the standard deviation which is simple to do .

It is almost the exact same formula . The only difference is that they have lnx instead of x and that there is an x in the denominator which mine doesn't have .

[/ QUOTE ]

Doesn't that just give you the pdf for the Normal Distribution?

PairTheBoard

[ApeAttack]

[ QUOTE ]
anyone?

the long normal distribution should not be undefined @ x=0

however the above equation is, right?

[/ QUOTE ]

Notice how there is an e^[-(lnx)^2] term.

As x-->0, this term --> e^[-(-inf)^2] = e^-inf = 0

Use L'Hopital's rule to see what this whole function converges to at x-->0

[HP]

yeah, what tripped me up was I thought the limit as x --> 0+ of exp(-(ln(x))^2) was 1. /images/graemlins/frown.gif

[BruceZ]

[ QUOTE ]
[ QUOTE ]
This is a typo !! Ln x should be replaced with x and the denominator should not have an x .
This is the correct formula .

f(x) = 1/[sqrt(2*pi)sigma]*e^[-(x-mu)^2/(2*sigma^2)]

where sigma =standard deviation
mu is the mean and x takes on the values from negative infinity to positive infinity . Think of f(x) as the y-value and naturally x is on the x-axis . We can evaluate the height of this function by plugging in the appropriate numbers .

In this case x takes on the mean mu , then all you have to do is compute the standard deviation which is simple to do .

It is almost the exact same formula . The only difference is that they have lnx instead of x and that there is an x in the denominator which mine doesn't have .

[/ QUOTE ]

Doesn't that just give you the pdf for the Normal Distribution?

PairTheBoard

[/ QUOTE ]

Yes, the original reference is correct for the LOG normal. Just as the sum of a large number of i.i.d. random variables has a distribution that converges to a normal distribution by the central limit theorem, a product of such random variables follows the log normal distribution.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
This is a typo !! Ln x should be replaced with x and the denominator should not have an x .
This is the correct formula .

f(x) = 1/[sqrt(2*pi)sigma]*e^[-(x-mu)^2/(2*sigma^2)]

where sigma =standard deviation
mu is the mean and x takes on the values from negative infinity to positive infinity . Think of f(x) as the y-value and naturally x is on the x-axis . We can evaluate the height of this function by plugging in the appropriate numbers .

In this case x takes on the mean mu , then all you have to do is compute the standard deviation which is simple to do .

It is almost the exact same formula . The only difference is that they have lnx instead of x and that there is an x in the denominator which mine doesn't have .

[/ QUOTE ]

Doesn't that just give you the pdf for the Normal Distribution?

PairTheBoard

[/ QUOTE ]

Yes, the original reference is correct for the LOG normal. Just as the sum of a large number of i.i.d. random variables has a distribution that converges to a normal distribution by the central limit theorem, a product of such random variables follows the log normal distribution.

[/ QUOTE ]

These would have to be iid random variables which only take on positive values though. Right?

PairTheBoard

[arahant]

[ QUOTE ]
okay I just checked this (http://mathworld.wolfram.com/LogNormalDistribution.html) website and it says the same. I will be in deep thought trying to understand why the function is 0 @ x = 0

any help appreciated..

edit: nevermind. exp(-infinity) is 0, not 1

all is well with the universe

[/ QUOTE ]

Yeah, I read this and would have replied, but I honestly figured you'd see the error in whatever you were thinking before I hit 'post'. Plus, I didn't know what your problem was with it exactly. You just said 'wtf'...next time, try being more explicit /images/graemlins/smile.gif