lognormal scaling?

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

Hi there,

In this paper

http://polymer.bu.edu/hes/articles/lgcmps99.pdf

on page 3 in fig 3b they show scaled probability versus scaled volatility.

They describe the scaling as log-normal scaling with

\sqrt(v) exp(a-v/4)P(VT) as a function of [ln(V(T)-a]/\sqrt{\pi v}, where a and v are the mean and the "width" on a logarithmic scale

I don't really understand what they mean here?

a is just the logarithmic mean
v I assume in the standard deviation?

So why are the observations not just re-scales as (V(T)-a)/v?

outermeasure · **Posted:** Wed, 16 May 2012 16:21:35 UTC

Bazman wrote:

Hi there,

In this paper

http://polymer.bu.edu/hes/articles/lgcmps99.pdf

on page 3 in fig 3b they show scaled probability versus scaled volatility.

They describe the scaling as log-normal scaling with

\sqrt(v) exp(a-v/4)P(VT) as a function of [ln(V(T)-a]/\sqrt{\pi v}, where a and v are the mean and the "width" on a logarithmic scale

I don't really understand what they mean here?

a is just the logarithmic mean
v I assume in the standard deviation?

So why are the observations not just re-scales as (V(T)-a)/v?

No! If you scale V_T

by $\dfrac{V_T-a}{\nu}$ , then what happens to $\log V_T$ ? Clearly that is not what you want.

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

You would get:

$log(V_Tscaled)=log(VT-a)-log(\nu)$

but that is not what they do, I believe (although I admit the instructions are not clear hence my confusion).

instead they use $log(V_Tscaled)=\frac{log(V_T)-a}{\sqrt{\pi *\nu}}$

where a is the natural log (not $\log_{10}$ ) of the expectation of the data and $\nu$ is the natural log of the standard deviation calculated as shown below:

http://www.chinarel.com/onlinebook/Acce ... bution.htm

where as shown they take the width to be 3 standard deviations (of the median rank values).

So I am still confused by several points:

1.) Why is the probability weighted as shown in the original article? It seems to give the correct results but I would like to know how it was derived?
2.) Why is the denominator of the volatility scaling $\sqrt{\pi\nu}$ and not just $\nu$ given that $\nu$ in the new article above is the standard deviation and not the variance? again it works when I treat $\nu=\sigma$ where $\sigma$ is as shown in the new article but I would like to understand why? + I have no idea why the $\sqrt{\pi}$ is there.
3.) Where does the formula for the calculation of the standard deviation on the log log graph in the new article come from? It works but why?
4.) When calculating the standard deviation in the article in the new link they use median rank as opposed to just the standard 3 deviations? Is it necessary to use the median rank? Again it seems to give the correct answer although the difference with just 3 standard deviations is small.
5.) in the last part they take the base case to be $\exp(-(\ln(X))^2)$ however this function is not defined for negative values? + shouldn't the equation be $\frac{\exp(-(\ln(X))^2)}{2\nu}$ ? I get round this be calculating the base case function on a domain on the same length as the rescaled range but starting at zero and then plot it against the rescaled range which seems to work. Just want to check that my workaround here is correct?

even if you only answer some of the questions above you would be helping me a lot!

outermeasure · **Posted:** Thu, 17 May 2012 15:59:19 UTC

Bazman wrote:

You would get:

$log(V_Tscaled)=log(VT-a)-log(\nu)$

but that is not what they do, I believe (although I admit the instructions are not clear hence my confusion).

instead they use $log(V_Tscaled)=\frac{log(V_T)-a}{\sqrt{\pi *\nu}}$

where a is the natural log (not $\log_{10}$ ) of the expectation of the data and $\nu$ is the natural log of the standard deviation calculated as shown below:

http://www.chinarel.com/onlinebook/Acce ... bution.htm

where as shown they take the width to be 3 standard deviations (of the median rank values).

So I am still confused by several points:

1.) Why is the probability weighted as shown in the original article? It seems to give the correct results but I would like to know how it was derived?
2.) Why is the denominator of the volatility scaling $\sqrt{\pi\nu}$ and not just $\nu$ given that $\nu$ in the new article above is the standard deviation and not the variance? again it works when I treat $\nu=\sigma$ where $\sigma$ is as shown in the new article but I would like to understand why? + I have no idea why the $\sqrt{\pi}$ is there.
3.) Where does the formula for the calculation of the standard deviation on the log log graph in the new article come from? It works but why?
4.) When calculating the standard deviation in the article in the new link they use median rank as opposed to just the standard 3 deviations? Is it necessary to use the median rank? Again it seems to give the correct answer although the difference with just 3 standard deviations is small.
5.) in the last part they take the base case to be $\exp(-(\ln(X))^2)$ however this function is not defined for negative values? + shouldn't the equation be $\frac{\exp(-(\ln(X))^2)}{2\nu}$ ? I get round this be calculating the base case function on a domain on the same length as the rescaled range but starting at zero and then plot it against the rescaled range which seems to work. Just want to check that my workaround here is correct?

even if you only answer some of the questions above you would be helping me a lot!

Well,

is not

, but the density function. Hence the $\sqrt{\nu}$ (recall that if $X\sim N(\mu,\sigma^2)$ , then e^X

has density $\dfrac{1}{x\sqrt{2\pi\sigma^2}}\exp\left[-\dfrac{(\log x-\mu)^2}{2\sigma^2}\right]$ .

The $\exp(a+\nu/4)$ suggests $a=\mu,\nu=2\sigma^2$ .

The $\sqrt{\pi}$ is indeed not necessary. But you need a $\sqrt{\pi}$ somewhere anyway when you work with normal distribution and friends.

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

OK I think I am somewhat closer.

Using the result for the change of variables of a distribution shown about half way down this page:

http://en.wikipedia.org/wiki/Probabilit ... y_function

If the probability density function of a random variable X is given as f_X(x)

, it is possible to calculate the probability density function of Y=g(x)

using the formula

$f_Y(y)= \left|\frac{d}{dy}(g^{-1}(y))\right|.f_X(g^{-1}(y))$

Thus the lognormal distibution with $\mu=0$ and $\nu=1$

$P(V_T)=\frac{1}{V_T \sqrt{2 \pi}} \exp{(-\frac{1}{2}(\ln(V_T)^2))}$

Taking the change of variable to be

$g(V_T_{trans})=\frac{\ln(V_T)-\mu}{\sqrt{\nu}}$

I have excluded the $\sqrt{\pi}$ term shown in the original paper from the denominator as I agree with outermeasure that it doesn't seem to fit

and taking the inverse

$g(V_T_{trans})^{-1}=\exp(V_T_{trans}\sqrt{\nu}+\mu)$

and the derivative is

$g'(V_T_{trans})^{-1}=\sqrt{\nu}\exp(V_T_{trans}\sqrt{\nu}+\mu)$

and so applying the change of variable formula above.

$P(V_T_{trans})=\sqrt{\nu}\exp(V_T_{trans}\sqrt{\nu}+\mu)P(V_T)$

where now the transformed variable $V_T_{trans}=\frac{\ln(V_T)-\mu}{\sqrt{\nu}}$ is used in P(V_T)

instead of V_T

This is close to the form given in the sheet apart from the scaling factor on the probability is $\exp(V_T_{trans}\sqrt{\nu}+\mu)$ instead of $\exp(\frac{\nu}{4}+\mu)$

Can you see how to I can get to the solution given in the original paper from here?

Bazman · **Joined:** Mon, 29 Jan 2007 13:48:38 UTC **Posts:** 537

ignore previous post

OK I think I am somewhat closer.

Using the result for the change of variables of a distribution shown about half way down this page:

http://en.wikipedia.org/wiki/Probabilit ... y_function

If the probability density function of a random variable X is given as f_X(x)

, it is possible to calculate the probability density function of Y=g(x)

using the formula

$f_Y(y)= \left|\frac{d}{dy}(g^{-1}(y))\right|.f_X(g^{-1}(y))$

Thus the lognormal distibution with params $\mu$ and $\sigma$

$P(V_T)=\frac{1}{V_T \sigma \sqrt{2 \pi}} \exp{(-\frac{1}{2}(\frac{\ln(V_T-\mu_{LN}}{\sigma})^2))}$

and we want to transform to

$P(V_T_{trans})=\exp[-ln(x)^2]$ which they describe as the lognormal form in which $\mu=0$ and $\nu=1$

Taking the change of variable to be

$g(V_T)=\frac{\ln(V_T)-\mu_{LN}}{\sqrt{\nu \pi}}$

and $\nu=\frac{\sigma^2}{2}$ this effectively cancels out the $\frac{1}{2}$ term in the original pdf when you take it out of the square root and place it in the numerator.

and taking the inverse

$g(V_T_{trans})^{-1}=\exp(V_T_{trans}\sqrt{\nu \pi}+\mu_{LN})$

and the derivative is

$g'(V_T_{trans})^{-1}=\sqrt{\nu \pi}\exp(V_T_{trans}\sqrt{\nu \pi}+\mu_{LN})$

and so applying the change of variable formula above.

$P(V_T_{trans})=\sqrt{\nu \pi}\exp(V_T_{trans}\sqrt{\nu \pi}+\mu_{LN})P(V_T)$

where now the transformed variable $V_T_{trans}=\frac{\ln(V_T)-\mu_{LN}}{\sqrt{\nu \pi}}$ is used in P(V_T)

instead of V_T

now

$\mu_{LN} =\exp(\mu+\frac{\nu}{4})$

and $\sqrt{\nu\pi}\exp(V_T_{trans}\sqrt{\nu}) =\sqrt{\pi \nu}V_T_{trans}$

this cancels with the $\sqrt{\pi}\sigma V_T$ term in the denomiator of the first fraction in the original pdf leaving

$\frac{1}{\sqrt{2}}\sqrt{\nu} \exp{(\mu+\frac{\nu}{4})}P(V_T)$

which is the solution we are looking for apart from the factor of $\frac{1}{\sqrt{2}}$ can you see where that might cancel?

So the formula looks correct but I am not getting correct scaling when I apply to data.

My question is:

1.) When the variables are plotted in a log-log plot (as in fig 3a of the original paper), can I take $\mu$ to be equal to the value of the peak probability P(V_T)

?
If so should I then take $\ln(mu)$ or $\log_10(\mu)$ ?
Using the $\log_10(\mu)$ figure seems to work much better giving nearly right answers? But intuitively it seems wrong? The only place $\log_10$ scale is used is in the log-log plot?

2.) Similarly they say that $\nu$ should be the width of the parabola on a log-log scale? I simply take this to mean the difference in the values of V_T

when at the maximum and minimum probabilities respectively. Is this correct?
If so and the similar to part 1 should I use this value expressed on a $\ln()$ or $\log_10()$ scale?

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lognormal scaling?

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