Need help with expected value of probability distribution

In summary, the expected value of a probability distribution is something like $dt$, but I'm not sure how to solve for it.
  • #1
andresthor
3
0

Homework Statement



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

Homework Equations



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

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:

[tex]
\begin{equation}
\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)
\end{equation}
[/tex]

But then I get nowhere.
 
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  • #2
For an integral of the form

[tex]\int xe^{-k(x-a)^2} dx[/tex]

I would write it as

[tex]\int (x-a)e^{-k(x-a)^2} dx+ a\int e^{-k(x-a)^2} dx[/tex]

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

[tex]\int_{-\infty}^\infty e^{-t^2}\, dt = \sqrt\pi[/tex]
 
  • #3
Ok. I gave it another shot but I don't seem to be able to come up with a reasonable solution.

[tex]
\begin{equation}
\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)/ =
\end{equation}
[/tex]
[tex]
\begin{equation}
\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
\end{equation}
[/tex]

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.
 
  • #4
Hint: The integrand of the left integral is an odd function of u. (And the dx should be du.)
 
  • #5
andresthor said:
Ok. I gave it another shot but I don't seem to be able to come up with a reasonable solution.

[tex]
\begin{equation}
\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)/ =
\end{equation}
[/tex]
[tex]
\begin{equation}
\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
\end{equation}
[/tex]


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:

[tex]v = \frac {u^2} {\sqrt{4Dt}}[/tex]

Or you could observe the integrand is an odd function :wink:
 
  • #6
Of course! It's an odd function!

Thanks for the help, really appreciated.
 

Related to Need help with expected value of probability distribution

What is expected value?

Expected value is a measure of the average outcome of a random variable. It is calculated by multiplying each possible outcome by its probability and adding all of the products together.

How do you calculate expected value?

To calculate expected value, you multiply each possible outcome by its probability and add all of the products together. This can also be represented by the formula E(x) = Σx * P(x), where x represents the outcome and P(x) represents the probability of that outcome.

Why is expected value important?

Expected value is important because it helps us make informed decisions and predict outcomes in situations with uncertainty. It can also be used to determine the fairness of a game or experiment.

What is the difference between expected value and actual value?

Expected value is a predicted average outcome based on probabilities, while actual value is the real outcome that occurs. Expected value may not always match the actual value, as it is based on probabilities and not guaranteed outcomes.

How does expected value relate to probability distributions?

Expected value is a measure of central tendency in a probability distribution. It represents the average outcome of a random variable and can be used to compare different probability distributions. It is also used to calculate other important measures such as variance and standard deviation.

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