Need help with expected value of probability distribution

[andresthor]

Homework Statement

I'm trying to find the expected value of a probability distribution.

Homework Equations

$$\int_{-\infty}^\infty xP(x,t) = \int_{-\infty}^\infty x \frac{1}{\sqrt{4\pi Dt}}e^{-\frac{(x-dt)^2}{4Dt}}dx$$

The Attempt at a Solution

I expect the value to be something like $dt$ but then again I might be way off.
I've tried some different substitutions but have had no luck. One example of what I've been trying:

$$\begin{equation}<br /> \frac{1}{\sqrt{4\pi Dt}}\int_{-\infty}^\infty x e^{-\frac{(x-dt)^2}{4Dt}}dx=\left/ u=x-dt\right/ = \frac{1}{\sqrt{4\pi Dt}}\int_{-\infty}^\infty (u+dt) e^{-\frac{u^2}{4Dt}}du = \frac{1}{\sqrt{4\pi Dt}}\left(\int_{-\infty}^\infty u e^{-\frac{u^2}{4Dt}}du + dt \int_{-\infty}^\infty e^{-\frac{u^2}{4Dt}}du\right)<br /> \end{equation}$$

But then I get nowhere.

 

[LCKurtz]

For an integral of the form

$$\int xe^{-k(x-a)^2} dx$$

I would write it as

$$\int (x-a)e^{-k(x-a)^2} dx+ a\int e^{-k(x-a)^2} dx$$

and try the substitution u = (x-a)² on the first one. Then you can use the fact that

$$\int_{-\infty}^\infty e^{-t^2}\, dt = \sqrt\pi$$

 

[andresthor]

Ok. I gave it another shot but I don't seem to be able to come up with a reasonable solution.

$$\begin{equation}<br /> \frac{1}{\sqrt{4\pi Dt}}\int_{-\infty}^\infty x e^{-\frac{(x-t)^2}{4Dt}} dx=\frac{1}{\sqrt{4\pi Dt}} \left( \int_{-\infty}^\infty (x-dt) e^{-\frac{(x-dt)^2}{4Dt}} dx - dt \int_{-\infty}^\infty e^{-\frac{(x-dt)^2}{4Dt}} dx \right) =/u=(x-dt)/ = <br /> \end{equation}$$
$$\begin{equation}<br /> \frac{1}{\sqrt{4\pi Dt}} \int_{-\infty}^\infty (u) e^{-\frac{u^2}{4Dt}} dx - \frac{1}{\sqrt{4\pi Dt}} dt \int_{-\infty}^\infty e^{-\frac{(x-dt)^2}{4Dt}} dx<br /> \end{equation}$$

Now I'm pretty sure the right one becomes dt*1, but the other one has me a bit confused. Does the left one equal pi times the constant even though I have 4Dt in the denominator under u?
I'm just completely lost here.

 

[vela]

Hint: The integrand of the left integral is an odd function of u. (And the dx should be du.)

 

[LCKurtz]

Ok. I gave it another shot but I don't seem to be able to come up with a reasonable solution.

$$\begin{equation}<br /> \frac{1}{\sqrt{4\pi Dt}}\int_{-\infty}^\infty x e^{-\frac{(x-t)^2}{4Dt}} dx=\frac{1}{\sqrt{4\pi Dt}} \left( \int_{-\infty}^\infty (x-dt) e^{-\frac{(x-dt)^2}{4Dt}} dx - dt \int_{-\infty}^\infty e^{-\frac{(x-dt)^2}{4Dt}} dx \right) =/u=(x-dt)/ = <br /> \end{equation}$$
$$\begin{equation}<br /> \frac{1}{\sqrt{4\pi Dt}} \int_{-\infty}^\infty (u) e^{-\frac{u^2}{4Dt}} dx - \frac{1}{\sqrt{4\pi Dt}} dt \int_{-\infty}^\infty e^{-\frac{(x-dt)^2}{4Dt}} dx<br /> \end{equation}$$


Now I'm pretty sure the right one becomes dt*1, but the other one has me a bit confused. Does the left one equal pi times the constant even though I have 4Dt in the denominator under u?
I'm just completely lost here.

In the integral on the left (which should have du instead of dx) try the substitution:

$$v = \frac {u^2} {\sqrt{4Dt}}$$

Or you could observe the integrand is an odd function

 

[andresthor]

Of course! It's an odd function!

Thanks for the help, really appreciated.