Dirac Delta: Normal -> Lognormal?

In summary, the paper discusses the transformation of a Dirac delta function into a lognormal distribution, exploring the mathematical implications and applications of this transition. It examines the properties of the delta function in relation to stochastic processes and the conditions under which such a transformation is valid, highlighting its relevance in various fields such as finance and physics. The authors provide insights into the underlying mechanisms driving this change and its potential impacts on modeling and analysis.
  • #1
Steve Zissou
64
2
TL;DR Summary
One way to think of the Dirac Delta "function" is the limit of a normal distribution as its standard deviation collapses to an infinitesimal. What if we start with a lognormal?
Hello shipmates,

Instead of imagining a Dirac Delta as the limit of a normal, like this:
$$ \delta\left ( x \right ) = \lim_{a \to 0}\frac{1}{|a|\sqrt{2\pi}}\exp\left [ -\left ( x/a \right )^2 \right ] $$
Could we say the same thing except starting with a lognormal, like this?
$$ \delta_{LN} \left ( x \right ) = \lim_{a \to 0}\frac{1}{|a|x\sqrt{2\pi}}\exp\left [ -\left ( \log{x}/a \right )^2 \right ] $$

Thanks!

Your pal,
Stevsie
 
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  • #2
Then what is physical dimension of x ? It seems x cannot be a physical variable in your formula.
 
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  • #3
anuttarasammyak said:
Then what is physical dimension of x ? It seems x cannot be a physical variable in your formula.
x is Specific Gravity. It has no dimensions.
 
  • #4
Thanks. I plot https://www.wolframalpha.com/input?...x*a)e^(-+log(x)^2/a^2)+for+a=0.1+and+a=0.001+
1721344970418.png
 
  • #5
This does not look to me like a delta function. δ(-x) is zero for positive x. Your function isn't even defined.
 
  • #6
Vanadium 50
Right. Good point.
 
  • #7
Vanadium 50
Do you think it seems legit if we say this:
$$ \delta_{LN}\left ( x-\mu \right )=\lim_{a \to 0^+}\frac{1}{|a|x\sqrt{2\pi}}\exp\left [ -\left ( \frac{\log{x}-\mu}{a} \right )^2 \right ] $$
...and restrict our line of thinking so that ## x>0 ## throughout a bigger problem. Does this seem legitimate to you?
Thanks
Stevsie
 
  • #8
Steve Zissou said:
restrict our line of thinking
Now we're getting into the question "how close is close enough?" I think that can't be answered in the abstract. It can only be answered only if we know how close is close enough in advance.
 
  • #9
Steve Zissou said:
TL;DR Summary: One way to think of the Dirac Delta "function" is the limit of a normal distribution as its standard deviation collapses to an infinitesimal. What if we start with a lognormal?

Could we say the same thing except starting with a lognormal, like this?
δLN(x)=lima→01|a|x2πexp⁡[−(log⁡x/a)2]

Thanks!
[tex]\delta_{LN}(x)=\frac{\delta ( \log x )}{x}=\frac{\delta(x-1)}{x}=\delta(x-1)[/tex]
defined in x>0. Plot in #4 shows it. BTW how do you design specific gravity x plays a role in the formula ?
 
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  • #10
Steve Zissou said:
TL;DR Summary: One way to think of the Dirac Delta "function" is the limit of a normal distribution as its standard deviation collapses to an infinitesimal. What if we start with a lognormal?
Yes, you can get a Dirac delta function (but not exactly the one you wrote down) by starting with a lognormal distribution ##\phi##:$$\phi\left(x>0\right)=\frac{1}{x\sigma\sqrt{2\pi}}\exp\left[-\frac{1}{2}\left(\frac{\ln x-\mu}{\sigma}\right)^{2}\right],\quad\phi\left(x\leq0\right)=0\tag{1}$$where ##\mu,\sigma## are the mean and standard deviation of the distribution. First integrate ##\phi## to find the cumulative distribution function ##\Phi##:$$\varPhi\left(x>0\right)=\intop_{0}^{x}\phi\left(x^{\prime}\right)dx^{\prime}=\frac{1}{2}\text{erfc}\left(\frac{\mu-\ln x}{\sigma\sqrt{2}}\right)\tag{2}$$in terms of the complementary error function. By the limiting properties of ##\text{erfc}\left(-z/(\sigma\sqrt{2})\right)##, eq.(2) reduces to the Heaviside step function ##\theta(z)## as the standard deviation goes to zero:$$
\
\lim_{\sigma\rightarrow0^{+}}\varPhi\left(x\right)=\lim_{\sigma\rightarrow0^{+}}\frac{1}{2}\text{erfc}\left(\frac{\mu-\ln x}{\sigma\sqrt{2}}\right)=\begin{cases}
\begin{array}{c}
1\\
0
\end{array} & \begin{array}{c}
\left(\ln x>\mu\right)\\
\left(\ln x<\mu\right)
\end{array}\end{cases}\equiv\theta\left(\ln x-\mu\right)
\tag{3}$$Now differentiate (3) with respect to ##\text{ln }x## obtain a Dirac delta function:$$\frac{d}{d\ln x}\left[\lim_{\sigma\rightarrow0^{+}}\varPhi\left(x\right)\right]=\frac{d}{d\ln x}\left[\theta\left(\ln x-\mu\right)\right]=\delta\left(\ln x-\mu\right)\tag{4}$$Then, (using physicists math!) assume that it's OK to exchange the order of the derivative and limit operations to get:$$\delta\left(\ln x-\mu\right)=\lim_{\sigma\rightarrow0^{+}}\frac{d}{d\ln x}\varPhi\left(x\right)=\lim_{\sigma\rightarrow0^{+}}x\frac{d}{dx}\varPhi\left(x\right)=\lim_{\sigma\rightarrow0^{+}}x\,\phi\left(x\right)\tag{5}$$or finally, in view of eq.(1):$$\delta\left(\ln x-\mu\right)=\lim_{\sigma\rightarrow0^{+}}\left[\frac{1}{\sigma\sqrt{2\pi}}\exp\left[-\frac{1}{2}\left(\frac{\ln x-\mu}{\sigma}\right)^{2}\right]\right]\tag{6}$$where ##x## is restricted to be positive.
 
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  • #11
renormalize said:
Yes, you can get a Dirac delta function (but not exactly the one you wrote down) by starting with a lognormal distribution ##\phi##:$$\phi\left(x>0\right)=\frac{1}{x\sigma\sqrt{2\pi}}\exp\left[-\frac{1}{2}\left(\frac{\ln x-\mu}{\sigma}\right)^{2}\right],\quad\phi\left(x\leq0\right)=0\tag{1}$$where ##\mu,\sigma## are the mean and standard deviation of the distribution. First integrate ##\phi## to find the cumulative distribution function ##\Phi##:$$\varPhi\left(x>0\right)=\intop_{0}^{x}\phi\left(x^{\prime}\right)dx^{\prime}=\frac{1}{2}\text{erfc}\left(\frac{\mu-\ln x}{\sigma\sqrt{2}}\right)\tag{2}$$in terms of the complementary error function. By the limiting properties of ##\text{erfc}\left(-z/(\sigma\sqrt{2})\right)##, eq.(2) reduces to the Heaviside step function ##\theta(z)## as the standard deviation goes to zero:$$
\
\lim_{\sigma\rightarrow0^{+}}\varPhi\left(x\right)=\lim_{\sigma\rightarrow0^{+}}\frac{1}{2}\text{erfc}\left(\frac{\mu-\ln x}{\sigma\sqrt{2}}\right)=\begin{cases}
\begin{array}{c}
1\\
0
\end{array} & \begin{array}{c}
\left(\ln x>\mu\right)\\
\left(\ln x<\mu\right)
\end{array}\end{cases}\equiv\theta\left(\ln x-\mu\right)
\tag{3}$$Now differentiate (3) with respect to ##\text{ln }x## obtain a Dirac delta function:$$\frac{d}{d\ln x}\left[\lim_{\sigma\rightarrow0^{+}}\varPhi\left(x\right)\right]=\frac{d}{d\ln x}\left[\theta\left(\ln x-\mu\right)\right]=\delta\left(\ln x-\mu\right)\tag{4}$$Then, (using physicists math!) assume that it's OK to exchange the order of the derivative and limit operations to get:$$\delta\left(\ln x-\mu\right)=\lim_{\sigma\rightarrow0^{+}}\frac{d}{d\ln x}\varPhi\left(x\right)=\lim_{\sigma\rightarrow0^{+}}x\frac{d}{dx}\varPhi\left(x\right)=\lim_{\sigma\rightarrow0^{+}}x\,\phi\left(x\right)\tag{5}$$or finally, in view of eq.(1):$$\delta\left(\ln x-\mu\right)=\lim_{\sigma\rightarrow0^{+}}\left[\frac{1}{\sigma\sqrt{2\pi}}\exp\left[-\frac{1}{2}\left(\frac{\ln x-\mu}{\sigma}\right)^{2}\right]\right]\tag{6}$$where ##x## is restricted to be positive.
Outstanding. Thank you.
 
  • #12
Thanks all.
 
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