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I know that the characteristic function $f(t, u) = \mathbb{E}(e^{iuX_t})$ of some random variable $X_t$ depending on $t \geq 0$ has to solve: $$f_t(t, u) = \left(iu - \frac{u^2}{2}\right) f(t, u) + u f_u(t, u) + \frac{u^2}{2}f_{uu}(t, u),$$ with the following boundary conditions: $f(0, u) = e^{iuy}$, $f(t, 0) = 1$ and $|f(t, u)| \leq 1$, for all $u \in \mathbb{R}$, $t \geq 0$.

I would like to obtain an explicit solution of this equation, if possible. It is not possible to use the separation of variables since the boundary conditions are not homogenous. An other idea is to put $f(t, u) = \exp(g(t, u))$, to obtain: $$g_t(t, u) = \left(iu - \frac{u^2}{2}\right) + u g_u(t, u) + \frac{u^2}{2}\left(g_u(t, u)^2 + g_{uu}(t, u)\right),$$ with the following boundary conditions: $g(0, u) = iuy$, $g(t, 0) = 0$ and $g(t, u) \leq 0$, for all $u \in \mathbb{R}$, $t \geq 0$.

Does anyone have an idea how to solve either equation? Thanks a lot!

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  • $\begingroup$ What is it about the initial and boundary conditions that prevents you from using separation of variables? $\endgroup$
    – MasterYoda
    Sep 18, 2018 at 1:50
  • $\begingroup$ If we set $f(t, u) = \alpha(t) \beta(u)$ for some functions $\alpha$, $\beta$, the condition $f(t, 0) = 1$, for all $t \geq 0$, would imply $\alpha(t)\beta(0) = 1$, for all $t \geq 0$, or equivalently (that $\beta(0) \neq 0$ and) that $\alpha$ is constant. $\endgroup$
    – kantadou
    Sep 18, 2018 at 8:09
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    $\begingroup$ In the condition $f(0,u)=e^{iuy}$ what is $y$ which appears here coming from nowhere ? $\endgroup$
    – JJacquelin
    Sep 23, 2018 at 8:55
  • $\begingroup$ Thanks, I forgot to mention that $y \in \mathbb{R}$ is a fixed constant (corresponding to the non-random value of $X_0$). $\endgroup$
    – kantadou
    Sep 24, 2018 at 8:21

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