Consider the following integral. Given fixed $x,y>0$, $$I(x,y):=\int_0^1\frac{x}{\sqrt{2\pi t^3}}\exp\left(-\frac{x^2}{2t}\right)\cdot\frac{1}{\sqrt{2\pi(1-t)}}\exp\left(-\frac{y^2}{2(1-t)}\right)~\mathrm dt.$$

I have reason to believe that $$I(x,y)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(x+y)^2}{2}\right),$$ but can't prove this.

The closest I got to a solution (though it's not much) is this: If $(W_t)_{t\geq0}$ is a standard Brownian motion and $$H_x:=\inf\{t\geq0:W_t=x\}$$ is the first hitting time of some point $x>0$, then, $$\mathbb P\{H_x=\mathrm dt\}=\frac{x}{\sqrt{2\pi t^3}}\exp\left(-\frac{x^2}{2t}\right).$$

By the strong Markov property, if $H_x$ and $H_y$ are independent Brownian hitting times, then $$H_x+H_y=_dH_{x+y},$$ where the above is an equality in distribution. Thus, it follows from a convolution argument that if $$I'(x,y):=\int_0^1\frac{x}{\sqrt{2\pi t^3}}\exp\left(-\frac{x^2}{2t}\right)\cdot\frac{y}{\sqrt{2\pi (1-t)^3}}\exp\left(-\frac{y^2}{2(1-t)}\right)~\mathrm dt,$$ then $$I'(x,y)=\mathbb P\{H_{x+y}=\mathrm d1\}=\frac{x+y}{\sqrt{2\pi}}\exp\left(-\frac{(x+y)^2}{2}\right).$$

So the integral $I(x,y)$ is somewhat similar, but instead of being a convolution of two hitting time densities, it is the convolution of one hitting time density together with a Gaussian density where the variance is integrated.

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    $\begingroup$ The lack of symmetry in $x$ and $y$ would seem to make a result that's symmetric in $x$ and $y$ unlikely. $\endgroup$ – StephenG Aug 2 '17 at 0:55
  • $\begingroup$ Looks like there is symmetry... $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(a+b-t) dt$, so if, $F(x,y) = \frac{\partial{I(x,y)}}{\partial y}$, then $F(x,y) = F(y,x)$ $\endgroup$ – Aryabhata Aug 2 '17 at 2:25

Using Aryabhata's comment as an inspiration, we see that, using my question's notation, $$\frac{\partial I(x,y)}{\partial y}=-I'(x,y)=-\frac{(x+y)}{\sqrt{2\pi}}\exp\left(-\frac{(x+y)^2}{2}\right).$$

Therefore, there is some function $f$ of $x$ such that $$I(x,y)=\int -I'(x,y)~\mathrm dy=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(x+y)^2}{2}\right)+f(x).$$

In order to show that $f(x)=0$, we need only prove that $$I(x,0)=\int_0^1\frac{x}{\sqrt{2\pi t^3}}\exp\left(-\frac{x^2}{2t}\right)\cdot\frac{1}{\sqrt{2\pi(1-t)}}~\mathrm dt=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right).$$

To see this, note that the indefinite integral (Mathematica) is given by $$\int\frac{x}{\sqrt{2\pi t^3}}\exp\left(-\frac{x^2}{2t}\right)\cdot\frac{1}{\sqrt{2\pi(1-t)}}~\mathrm dt=-\frac{e^{-\frac{x^2}{2}} \text{erf}\left(\frac{x}{\sqrt{2} \sqrt{\frac{1}{1-t}-1}}\right)}{\sqrt{2 \pi }},$$ where $$\mathrm{erf}(z):=\frac{2}{\sqrt{\pi}}\int_0^ze^{-x^2}~\mathrm dx$$ is the error function. At this point, we simply use the fundamental theorem of calculus.


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