Using the Heat Kernel to Recover Initial Data in Black–Scholes

Summary

The document considers a Black–Scholes equation transformed into the heat equation and asks why its integral solution approaches the original function as time tends to zero. The proposed first step is a change of variables that rescales the integration coordinate by the square root of time. This rewrites the expression in terms of a Gaussian kernel centered on the point of interest, which helps show how the integral samples values near that point as the kernel narrows.

The response also suggests interchanging the limit and integral. That step requires appropriate mathematical justification; continuity alone may not be sufficient over the whole real line without further conditions on the function or a separate argument controlling the tails. The excerpt gives a direction rather than a complete proof, and it does not discuss the assumptions needed for the limit or connect the result to option valuation beyond the Black–Scholes transformation.

Key ideas

• A variable change scaled by the square root of time turns the integral into a Gaussian-kernel expression.
• As time approaches zero, the kernel concentrates near the point where the initial function is evaluated.
• The response recommends exchanging the limit and integral but does not provide a justification.
• A rigorous limit argument may require assumptions beyond continuity to control contributions far from the evaluation point.

Tags

# Heat/Diffusion Equation


# Heat/Diffusion Equation












I am working on a problem where I have successfully reduced a version of Black Scholes to the Heat Equation and then shown the solution to be:

$$u(x,t)=\frac{1}{2\sqrt{t\pi}}\int_{-\infty}^\infty{f(\xi)e^{-\frac{(x-\xi)^2}{4t}}}d\xi$$

I now need to show that if $f(x)$ is continuous then $$\lim_{t\rightarrow 0+}u(x,t)=f(x)$$

Further, there is a tip that a change of variables of $p=\frac{(\xi-x)}{2\sqrt{t}}$ may help.

I think that I need to do some integration by parts, show that some part of the integration go to zero as $t\rightarrow0+$ which will cancel out and then by linearity I can say that as u(x,t) is a solution and then $f(x)$ must also be a solution. I am just missing the first step and hoped someone can give me a nudge!

Thanks for any help!

## Answer by Gordon (score 5)

https://quant.stackexchange.com/a/14215

What you need to do is to first make a variable change such as $u = \frac{x-\xi}{2\sqrt{t}}$. Then change the order of the limit and the integral.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.