Property of gaussian integrals 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!
 A: 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.
A: 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!
