Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Apologies if this has been asked before... I came across the following relation:

if $$P(x_2, t_2 \mid x_1, t_1) = \frac{1}{\sqrt{2\pi\sigma^2(t_2-t_1)}}e^{-\frac{(x_2-x_1)^2}{2\sigma^2(t_2-t_1)}}$$ ($t_2 > t_1$) then we have the following identity, with $T>0$: $$P(x_1, T \mid x_0, 0)\delta(x_1-x_2)-P(x_1, T \mid x_0, 0)P(x_2, T \mid x_0, 0)$$ $$=\sigma^2 \int_0^T\int_x P(x, t \mid x_0, 0)\frac{\partial P(x_1, T \mid x, t)}{\partial x}\frac{\partial P(x_2, T \mid x, t)}{\partial x}dxdt$$

I tried several things but none came close to a beginning of a solution...

Thanks a lot!

share|cite|improve this question

This is only a partial answer, where I assume that $x_1 \neq x_2$, so that the first term with Dirac's delta function disappears. I'm not sure of what the correct interpretation should be when $x_1 =x_2$, but computations with Maple indicate that the double integral is divergent then, and I assume the first term $P(x_1,T \mid x_0,0)\delta(x_1-x_2)$ tells how it diverges. But from now on, $x_1\neq x_2$ is assumed.

We will need the three integrals $$\int_{-\infty}^\infty {\rm e}^{-\beta u^2}{\rm d}u=\sqrt{\pi/\beta}$$ $$\int_{-\infty}^\infty u{\rm e}^{-\beta u^2}{\rm d}u=0$$ $$\int_{-\infty}^\infty u^2{\rm e}^{-\beta u^2}{\rm d}u={1\over 2}\sqrt{\pi/\beta^3}$$ which we will use with $u=x-{b\over a}$ and $\beta ={a\over 2\sigma^2}$, where $a$ and $b$ are determined below.

Multiplying together the three factors in the inner integral, we get the integrand $$A=(2\pi)^{-3/2}\sigma^{-7}{(x-x_1)(x-x_2) \over t^{1/2}(T-t)^3}{\rm exp}\Bigl(-{1\over 2\sigma^2}(ax^2-2bx+c)\Bigr)$$ where $$a={1\over t}+{1\over T-t}+{1\over T-t}$$ $$b={x_0\over t}+{x_1\over T-t}+{x_2\over T-t}$$ $$c={x_0^2\over t}+{x_1^2\over T-t}+{x_2^2\over T-t}$$ We will integrate this with respect to $x$. First complete the square to get $$ax^2-2bx+c=a\Bigl(x-{b\over a}\Bigr)^2+{ac-b^2\over a}$$ Here $${ac-b^2 \over a}={(T-t)(x_1-x_0)^2+(T-t)(x_2-x_0)^2+t(x_2-x_1)^2 \over T^2-t^2}$$ We also need to modify the factor in front of the exponential function in $A$, by finding $d$ and $e$ such that $(x-x_1)(x-x_2)=(x-{b\over a})^2+d(x-{b\over a})+e$. We won't need the exact value of $d$, and evaluating at $x={b\over a}$ we see that $$e=\Bigl({b\over a}-x_1\Bigr)\Bigl({b\over a}-x_2\Bigr)$$ $$={1 \over (T+t)^2} \bigl((T-t)(x_0-x_1)+t(x_2-x_1)\bigr)\bigl((T-t)(x_0-x_2)+t(x_1-x_2)\bigr)$$

We are now ready to compute the inner integral, using the three integral formulas mentioned initially together with the substitution $u=x-{b\over a}$. $$\int_{-\infty}^\infty A\,{\rm d}x={(2\pi)^{-3/2}\sigma^{-7}\over t^{1/2}(T-t)^3}\int_{-\infty}^\infty(u^2+du+e){\rm exp}\Bigl(-{1\over 2\sigma^2}\Bigl(au^2+{ac-b^2\over a}\Bigr)\Bigr){\rm d}u$$ $$={(2\pi)^{-3/2}\sigma^{-7}\over t^{1/2}(T-t)^3}\Bigl({1\over 2}\sqrt{{\pi \over (a/2\sigma^2)^3}}+0+e\sqrt{{\pi\over a/2\sigma^2}}\Bigr){\rm exp}\Bigl(-{1\over 2\sigma^2}\Bigl({ac-b^2\over a}\Bigr)\Bigr)$$ $$=(2\pi)^{-1}\sigma^{-4}\Bigl({t \over (T^2-t^2)^{3/2}}+\sigma^{-2}{B\over (T^2-t^2)^{5/2}}\Bigr){\rm exp}\Bigl(-{1\over 2\sigma^2}\Bigl({ac-b^2\over a}\Bigr)\Bigr)$$ Here $B$ in the last expression is given by $$B=\bigl((T-t)(x_0-x_1)+t(x_2-x_1)\bigr)\bigl((T-t)(x_0-x_2)+t(x_1-x_2)\bigr)$$ and we also have a nasty expression for ${ac-b^2\over a}$ to substitute, which is given above.

Now it's time for a miracle! We want to integrate this last expression with respect to $t$, and there is an explicit primitive function, given by $$F(t)=(2\pi)^{-1}\sigma^{-4}{{\rm exp}\Bigl(-{1\over 2\sigma^2}\Bigl({ac-b^2\over a}\Bigr)\Bigr) \over (T^2-t^2)^{1/2}}$$ Here $F(t)\to 0$ as $t\to T$, and hence $$\int_0^T\int_{-\infty}^\infty A \,{\rm d}x\,{\rm d}t=-F(0)=-(2\pi)^{-1}\sigma^{-4}{1\over T}{\rm exp}\Bigl(-{1\over 2\sigma^2}\Bigl({(x_1-x_0)^2\over T}+{(x_2-x_0)^2 \over T}\Bigr)$$ Multiplying by $\sigma^2$, this finally becomes $$-P(x_1,T\mid x_0,0)P(x_2,T\mid x_0,0)$$ as desired.

share|cite|improve this answer
Ok, that's impressive... I tried a direct calculation with no luck. However like you say with this method I fail to see how the dirac can pop up... Thanks! – mica_t Aug 9 '12 at 13:14
There should be a slick way of doing this, but sometimes brute force can feel quite satisfying too :-) – Per Manne Aug 9 '12 at 14:40
Well also I think there should be a square root around $(2\pi)^{-1} \frac{1}{T}$... tonight I'll try to redo your integration (it must be working this way) – mica_t Aug 9 '12 at 17:08

Ok, I was wrong obviously, PerManne's result is correct it took me a long time and quite a bit of paper to finally re-do everything.

For the Dirac part it is actually easier than I thought, just pick any function nice enough $g(x_1, x_2)$ and try to compute $I(t) = \int_{x_1}\int_{x_2}g(x_1, x_2)F(t, x_1, x_2)dx_1 dx_2$ where $F$ is defined as above, when $t \to T$.

Simple calculations lead to

$$I(t)=\frac{1}{\sigma^2} \int_{x_1}\int_{x_2} \frac{g(x_1, x_2)}{\sqrt{2\pi \sigma^2 t}} e^{-\frac{(x_1-x_0)^2+(x_2-x_0)^2}{2\sigma^2(T+t)}} \frac{e^{-\frac{(x_1-x_2)^2}{2\hat{\sigma}^2}}}{\sqrt{2\pi \hat{\sigma}^2}}dx_2dx_1$$ $$= \frac{1}{\sigma^2} \int_{x_1}\int_{x_2} h(t, x_1, x_2) \frac{e^{-\frac{(x_1-x_2)^2}{2\hat{\sigma}^2}}}{\sqrt{2\pi \hat{\sigma}^2}}dx_2dx_1$$

with $\hat{\sigma} = \sigma \sqrt{\frac{T^2-t^2}{t}}$, $\hat{\sigma} \to 0$ when $t \to T$.

Then the Dirac term comes from the second exponential since $\frac{e^{-\frac{(x_1-x_2)^2}{2\hat{\sigma}^2}}}{\sqrt{2\pi \hat{\sigma}^2}} \to +\infty$ when $x_1 = x_2$ and $t \to T$ however $\int_{x_2}\frac{e^{-\frac{(x_1-x_2)^2}{2\hat{\sigma}^2}}}{\sqrt{2\pi \hat{\sigma}^2}}dx_2 = 1$

So finally we get when $t \to T$, $$I(T) \to \frac{1}{\sigma^2}\int_{x_1} h(x_1, x_1, T)dx_1$$ with $$h(x_1, x_1, T) = g(x_1, x_1)P(x_1, T \mid x_0, 0)$$

(there might be a little more to say to prove this but this works)

And I guess we can write it like $$I(T) \to \frac{1}{\sigma^2}\int_{x_1}\int_{x_2} g(x_1, x_2)P(x_1, T \mid x_0, 0) \delta(x_2-x_1) dx_ 1dx_2$$ which is, without the integrals and the function $g$, what I asked.

Thanks a lot PerManne!

share|cite|improve this answer
You're welcome! Btw, it seems you have created two accounts on math.stackexchange. If you will be using this site more, you may want to merge them. – Per Manne Aug 13 '12 at 13:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.