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My understanding of Lecture #33, 34: The Characteristic Function for a Diffusion:


As an alternative to directly computing the characteristic function of a random variable $X_t$ in a stochastic process $\{X_t\}_{t \in [0,T]}$, we can solve a (boundary?) value problem, whose PDE has parameters are given by the dynamics of the stochastic process, and then conclude by Feynman-Kac that it is the characteristic function of said random variable.

An example is Arithmetic Brownian Motion:

If we solve

$$ \frac{\partial f}{\partial t} + \mu \frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2 \frac{\partial^2 f}{\partial x^2} = 0, x \in \mathbb R, t \in [0,T]$$ $$f(T,x) = e^{i \theta x} \tag{1}$$

then we get a function $f(t,x)$ s.t. $f(0,x)$ is the characteristic function of $X_t$ where $$dX_t = \sigma dW_t + \mu dt$$


So what does this mean for the (boundary?) value problem

$$ \frac{\partial f}{\partial t} + \mu x \frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2 x^2\frac{\partial^2 f}{\partial x^2} = 0, x \in \mathbb R, t \in [0,T]$$ $$f(T,x) = e^{i \theta x} \tag{2}$$

?

My guess is that solution of $(2)$, $f(t,x)$, will be s.t. $f(0,x)$ is the characteristic function of $X_t$ where $$dX_t = \sigma X_t dW_t + \mu X_t dt$$ i.e. $X_t$ is Geometric Brownian Motion and hence is lognormal, the distribution of which doesn't have a characteristic function.

So $(2)$ has no solution then?

I'm looking for an answer like

'We don't expect $(2)$ to have a solution. This can be proven through (some PDE things).'

or

'While we don't expect $(2)$ to have a solution, it actually does because (some PDE things), but then (some PDE things).'

So the (some PDE things) may or may not prove lognormal distribution doesn't have a characteristic function, but I'm looking more for consistency e.g. because

  1. Lognormal doesn't have a characteristic function
  2. For any random variable, its characteristic function is supposed to be able to be computed by solving a PDE
  3. of Feynman-Kac

said PDE must have no solution.

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    $\begingroup$ ? Any probability distribution has characteristic function... $\endgroup$ – m7e Jan 11 '17 at 18:53
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    $\begingroup$ The value at $t$ of the characteristic function of a random variable $X$ is the expected value of $e^{itX}$. Since (for real $t$), $|e^{itX}|=1$, and any probability measure is finite, $e^{itX(\omega)}$ is always integrable. What the wiki mentions is that for lognormal $X$, this function cannot be extended to complex $t$-s, in particular, it is not analytic. $\endgroup$ – m7e Jan 12 '17 at 9:18
  • $\begingroup$ @m7e My new understanding: 1. Lognormal does have a characteristic function 2. It has no closed form 3. The value problem solution doesn't have an Ansatz directly related to its terminal condition as with the value problems for ABM and OU 4. The value problem solution rather has a series solution. 5. Such series solution gives the characteristic function of lognormal for $t=0$. Is that right? $\endgroup$ – BCLC Mar 21 '18 at 15:35

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