Taylor Series and Transition Density Functions

See my Github repo for full details.

https://github.com/eightsmile/cqf

1. Trinomial Random Walk

2. Transition Probability Density Function

The transition probability density function, $p(y,t;y’,t’)$, is defined by,

$$Prob(a<y^{\prime }<b,attime{t}^{\prime }|yattimet)={\int }_a^{b}p(yet;y^{\prime },t^{\prime })d{y}^{\prime }$$

In words this is “the probability that the random variable y ′ lies between a and b at time t ′ in the future, given that it started out with value y at time t.”

Think of y and t as being current values with y ′ and t ′ being future values. The transition probability density function can be used to answer the question,

“What is the probability of the variable y ′ being in a specified range at time t ′ in the future given that it started out with value y at time t?”

Our Goal is to find the transition probability p.d.f., and so we find the relationship between $p(y,t;y’,t’)$, and $p(y,t;y’,t’-\delta t)$,

3. From the Trinomial model to the Transition Probability Density function

The variable y can either rise, fall or take the same value after a time step δt. These movements have certain probabilities associated with them.

We are going to assume that the probability of a rise and a fall are both the same, $\alpha<\frac{1}{2}$ . (But, of course, this can be generalized. Why would we want to generalize this?)

3.1 The Forward Equation

Given ${y,t}$, or says ${y,t}$ the current and previous. ${y’,t’}$ are variate in the future time.

The probability of being at $y’$ at time $t’$ is related to the probabilities of being at the previous three values and moving in the right direction:

$$p(y,t;y^{\prime },t^{\prime })=\alpha p(y,t;y^{\prime }+\delta y,t^{\prime }-\delta t)+(1-2\alpha )p(y,t;y^{\prime },t^{\prime }-\delta t)+\alpha p(y,t;y^{\prime }-\delta y,t^{\prime }-\delta t)$$

Given ${y,t}$, we find relationship between ${y’,t’}$ and ${y’\pm \delta y,t’-\delta t}$ that is y’ and t’ a bit time previously.

Remember, our goal is to find a solution of $p(.)$, we try to solve the above equation.

3.2 Taylor Series Expansion

We expand each term of the equation.

$$p(y,t;y^{\prime },t^{\prime })=\alpha p(y,t;y^{\prime }+\delta y,t^{\prime }-\delta t)+(1-2\alpha )p(y,t;y^{\prime },t^{\prime }-\delta t)+\alpha p(y,t;y^{\prime }-\delta y,t^{\prime }-\delta t)$$

Why we do that? Because there are too many variables in it, hard to solve it. We have to reduce the dimension.

$$p(y,t;y^{\prime }+\delta y,t^{\prime }-\delta t)\approx p(y,t;y^{\prime },t)–\delta t\frac{\partial p}{\partial t^{\prime }}+\delta y\frac{\partial p}{\partial y^{\prime }}+\frac{1}{2}\delta y^2\frac{{\partial }^{2}p}{\partial y^{\prime 2}}+O(\frac{{\partial }^{2}p}{\partial t^{\prime 2}})$$

$$p(y,t;y^{\prime },t^{\prime }-\delta t)\approx p(y,t;y^{\prime },t)–\delta t\frac{\partial p}{\partial t^{\prime }}+O(\frac{{\partial }^{2}p}{\partial t^{\prime 2}})$$

$$p(y,t;y^{\prime }-\delta y,t^{\prime }-\delta t)\approx p(y,t;y^{\prime },t)–\delta t\frac{\partial p}{\partial t^{\prime }}-\delta y\frac{\partial p}{\partial y^{\prime }}+\frac{1}{2}\delta y^2\frac{{\partial }^{2}p}{\partial y^{\prime 2}}+O(\frac{{\partial }^{2}p}{\partial t^{\prime 2}})$$

Plug them back into that equation, and after cancel out terms repeated we would left with,

$$\frac{\partial p}{\partial t^{\prime }}=\alpha \frac{\delta y^2}{\delta t}\frac{{\partial }^{2}p}{\partial y^{\prime 2}}+O(\frac{{\partial }^{2}p}{\partial t^{\prime 2}})$$

We drop those derivative terms with order greater and equal than $O(\frac{\partial^2 p}{\partial t’^2})$.

$$\frac{\partial p}{\partial t^{\prime }}=\alpha \frac{\delta y^2}{\delta t}\frac{{\partial }^{2}p}{\partial y^{\prime 2}}$$

In the RHS, we focus on $\alpha \frac{\delta y^2}{\delta t}$, firstly. The denominator and numerator have to be in the same order to make that term definite. Or, say $\delta y \sim O(\sqrt{\delta t})$.

We thus let $c^2 = \alpha \frac{\delta y^2}{\delta t}$

$$\frac{\partial p}{\partial t^{\prime }}=c^2\frac{{\partial }^{2}p}{\partial y^{\prime 2}}$$

The above equation is also named Fokker–Planck or forward Kolmogorov equation.

Now, we have a partial differential equation. Solve it, we can get the form of $p$.

3.3 Backward Equation works similar.

$$p(y,t;y^{\prime },t^{\prime })=\alpha p(y+\delta y,t+\delta t;y^{\prime },t^{\prime })+(1-2\alpha )p(y,t+\delta t;y^{\prime },t^{\prime })+\alpha p(y-\delta y,t+\delta t;y^{\prime },t^{\prime })$$

the dimension-reduced result is the blow, and it is called the backward Kolmogorov equation.

$$\frac{\partial p}{\partial t^{\prime }}+c^2\frac{{\partial }^{2}p}{\partial y^{\prime 2}}=0$$

4. Solve the Forward Kolmogorov Equation

We will solve for $p$ right now! However, we will solve it by assuming similarity solution.

$$\frac{\partial p}{\partial t^{\prime }}=c^2\frac{{\partial }^{2}p}{\partial y^{\prime 2}}$$

This equation has an infinite number of solutions. It has different solutions for different initial conditions and different boundary conditions. We need only a special solution here. The detailed process of finding that solution is showing as the following,

4. 1 Assume a Solution Form

$$p=t^{\prime a}f(\frac{y^{\prime }}{t^{\prime b}})=t^{\prime a}f(\xi )$$

$$\xi =\frac{y^{\prime }}{t^{\prime b}}$$

, where $a$, and $b$ are indefinite variables.

Again, don’t ask why it is in this form, because it is a special solution!

4.2 Derivation

$$\frac{\partial p}{\partial y^{\prime }}=t^{\prime a-b}\frac{df}{d\xi }$$

$$\frac{{\partial }^{2}p}{\partial y^{\prime 2}}=t^{\prime a-2b}\frac{d^{2}f}{d{\xi }^2}$$

$$\frac{\partial p}{\partial t^{\prime }}=a{t}^{\prime a-1}f(\xi )+b{y}^{\prime }{t}^{\prime a-b-1}\frac{df}{d\xi }$$

Substitue back into the forward Kolmogorov equation (remember $y’ = t’^b \xi$), we get,

$$af(\xi )–b\xi \frac{df}{d\xi }=c^2{t}^{\prime -2b+1}\frac{d^{2}f}{d{\xi }^2}$$

4.3 Choose $b$

As we need the RHS to be independent of $t’$, we could choose the value of $b=\frac{1}{2}$, to let the $t’$ has a power of 0. Why we do that? Because we aim to reduce the partial differential equation to be a ordinary differential equation, in which the only variable is $\xi$, and $t’$ disappear.

By assuming the special form of $p$, and letting $b= 1/2$, our forward Kolmogorov becomes,

$$af(\xi )–\frac{1}{2}\xi \frac{df}{d\xi }=c^2\frac{d^{2}f}{d{\xi }^2}$$

$$p=t^{\prime a}f(\frac{y^{\prime }}{\sqrt{t^{\prime }}})=t^{\prime a}f(\xi )$$

$$\xi =\frac{y^{\prime }}{\sqrt{t^{\prime }}}$$

4.4 Choose $a$

$$p=t^{\prime a}f(\frac{y^{\prime }}{\sqrt{t^{\prime }}})$$

We know that $p$ is the transition p.d.f., its integral must be equal to ‘1’. $t’$ is independent by the definition of random walk behaviour, so we do only integrate $p$, w.r.t. $y’$.

$${\int }_{\mathbb{R}}pd{y}^{\prime }={\int }_{\mathbb{R}}{t}^{\prime a}f(\frac{y^{\prime }}{\sqrt{t^{\prime }}})d{y}^{\prime }=1$$

$${\int }_{\mathbb{R}}{t}^{\prime a}f(\frac{y^{\prime }}{\sqrt{t^{\prime }}})d{y}^{\prime }=1$$

, by replace $x = \frac{y’}{\sqrt{t’}}$,

$${\int }_{\mathbb{R}}{t}^{\prime a+1/2}f(x)dx=t^{\prime a+1/2}{\int }_{\mathbb{R}}f(x)dx=1$$

$t^{\prime }$ is independent, so the above equation would be equal to ‘1’ regardless the power of $t^{\prime }$. Thus, $a=-\frac{1}{2}$ for sure.

Also, we get $\int_{\mathbb{R}} f(x)\ dx= 1$.

4.5 Integrate! Solve it!

By assuming the special form of $p$, and letting $a=-1/2$, $b=1/2$, we get,

$$-\frac{1}{2}f(\xi )–\frac{1}{2}\xi \frac{df}{d\xi }=c^2\frac{d^{2}f}{d{\xi }^2}$$

$$p=\frac{1}{\sqrt{t^{\prime }}}f(\frac{y^{\prime }}{\sqrt{t^{\prime }}})=\frac{1}{\sqrt{t^{\prime }}}f(\xi )$$

$$\xi =\frac{y^{\prime }}{\sqrt{t^{\prime }}}$$

The forward Kolmogorov equation becomes,

$$-\frac{1}{2}(f(\xi )–\xi \frac{df}{d\xi })=c^2\frac{d^{2}f}{d{\xi }^2}$$

$$-\frac{1}{2}(\frac{d\xi f(\xi )}{d\xi })=c^2\frac{d^{2}f}{d{\xi }^2}$$

, as $f(\xi) – \xi \frac{df}{d\xi} = \frac{d \xi f(\xi)}{d \xi}$.

Integrate 1st Time

$$-\frac{1}{2}\xi f(\xi )=c^2\frac{df}{d\xi }+constant$$

There’s an arbitrary constant of integration that could go in here but for the answer we want this is zero. We need only a special solution, so we can set that arbitrary constant term be zero.

So, the eq could be rewritten as,

$$-\frac{1}{2c^2}\xi d\xi =\frac{1}{f(\xi )}df$$

Integrate 2nd Time

$$lnf(\xi )=-\frac{{\xi }^2}{4c^2}+C$$

Take exponential, $f(\xi) = e^C e^{-\frac{\xi^2}{4c^2}} = A e^{-\frac{\xi^2}{4c^2}}$ .

Find $A$

The Last Step here is to find the exact value of $A$. $A$ is chosen such that the integral of $f$ is one.

$${\int }_{\mathbb{R}}f(\xi )d\xi =1$$

$${\int }_{\mathbb{R}}A{e}^{-\frac{{\xi }^2}{4c^2}}d\xi =2cA{\int }_{\mathbb{R}}{e}^{-\frac{{\xi }^2}{4c^2}}d(\frac{\xi }{2c})=1$$

$$2cA\sqrt{\pi }=1$$

, so we get $A = \frac{1}{2c\sqrt{\pi}}$

Plug $f(\xi), a, b, A$ back into $p = t^a f(\xi)$.

$$p(y^{\prime })=\frac{1}{2c\sqrt{\pi t^{\prime }}}{e}^{-\frac{{\xi }^2}{4c^2}}=\frac{1}{2c\sqrt{\pi t^{\prime }}}{e}^{-\frac{y^{\prime 2}}{4c^2{t}^{\prime }}}$$

$p(.)$ now is normal like distributed.

$$N(x)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}$$

So, we may say $\mu_{y’}=0$, and $\sigma^2_{y’}=2c^2t’$. Or, $y’ \sim N(0, 2c^2t’)$.

5. Summary

$$p(y^{\prime })=\frac{1}{2c\sqrt{\pi t^{\prime }}}{e}^{-\frac{{\xi }^2}{4c^2}}=\frac{1}{2c\sqrt{\pi t^{\prime }}}{e}^{-\frac{y^{\prime 2}}{4c^2{t}^{\prime }}}$$

Finally, we solved the transition probability density function $p(.)$. By assuming the forward or backward type of trinomial model, we find a partial differential relationship. Then, assuming a special form of $p(.)$ by similarity method, we solve it.

The meaning is that $p(y’)=\frac{1}{2c\sqrt{\pi \ t’}}e^{-\frac{\xi^2}{4c^2}} =\frac{1}{2c\sqrt{\pi \ t’}}e^{-\frac{y’^2}{4c^2t’}}$ is one of the transition probability density function that can satisfy the trinomial random walk.

Also, we find that $p(.)$ is normally liked distributed.