Dirac Delta: Normal -> Lognormal?

[Steve Zissou]

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

 

[anuttarasammyak]

Then what is physical dimension of x ? It seems x cannot be a physical variable in your formula.

 

[Steve Zissou]

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.

 

[anuttarasammyak]

Thanks. I plot https://www.wolframalpha.com/input?...x*a)e^(-+log(x)^2/a^2)+for+a=0.1+and+a=0.001+

 

[Vanadium 50]

This does not look to me like a delta function. δ(-x) is zero for positive x. Your function isn't even defined.

 

[Steve Zissou]

Vanadium 50
Right. Good point.

 

[Steve Zissou]

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

 

[Vanadium 50]

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.

 

[anuttarasammyak]

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!

$$\delta_{LN}(x)=\frac{\delta ( \log x )}{x}=\frac{\delta(x-1)}{x}=\delta(x-1)$$
defined in x>0. Plot in #4 shows it. BTW how do you design specific gravity x plays a role in the formula ?

 

[renormalize]

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.

 

[Steve Zissou]

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.

 

[Steve Zissou]

Thanks all.