I am having a issue with the wording of this question.

Find the probability of the following.

The velocity $v$ of a randomly selected particle, whose distribution obeys the probability density function

$$P(\frac{v}{v_{avg}})=exp(-\frac{v}{v_{avg}})$$ will lie between 0 and $2_{v_{avg}}$ where $v_{avg}$ is the average velocity.

I have been given the density function so I just integrated between the limits to find the probability between the given points.

which came to be $P(0\leq v \leq 2v_{avg})=0.86 v_{avg}$

However I am not happy with this as to me the probability can change given a different average velocity which dose not make sense.

I have looked at various probability distribution information in a text book am I using by John E.Frenuds and I cant see to find anything that jumps out then only thing that did is the use of 'probability density' where they define a given function say

$$f(x)=\left\{\begin{matrix} kx^2 >0 \\0 \: everywhere \: else \end{matrix}\right.$$

So then it get normalized and integrated between given limits, which make me think the above equation given is maybe a 'probability density function' as been used in the book.

Could someone maybe explain if I have use the correct method and if not, expand where I have gone wrong.

  • 2
    $\begingroup$ Note that the probability density you're given is the probability density of a variable $v/v_{avg}$, not just $v$. So you need to calculate $P(0<v/v_{avg}<2) = \int_0^2 e^{-x} dx$. $\endgroup$ – Adam Latosiński Apr 18 at 11:41
  • 1
    $\begingroup$ Please notice that your distribution P() is not a probability distribution function (PDF) of $v$ as it is not normalized. The PDF is rather $f(v)=\frac{1}{v_{avg}} Exp\left(- \frac{v}{v_{avg}}\right)$. The average of $v$ is then $<v> = \int_0^\infty v f(v) \,dv = v_{avg}$, the variance is $<(v-v_{avg})^2>= v_{avg}^2$. $\endgroup$ – Dr. Wolfgang Hintze Apr 18 at 12:25

The result of your calculation is correct, but the result itself requires the normalization factor $a=v_{avg}$, since $$\int_0^{\infty}e^{-\frac{t}{a}}\;dt = a$$

The probability is then independent of the average velocity because of the specific properties of the exponential distribution:

You have in general $$\frac {1}{a}\int_0^{2a}e^{-\frac{t}{a}}\;dt= \int_0^{2}e^{-x}\;dx = 1-\frac{1}{e^2}$$

With other probability distributions you may not see such a result in general.

  • $\begingroup$ In general $e^{-t/a}$ isn't a probability density, because it's not normalized. And if you do normalize it, the dependence of the final result on $a$ vanishes. $\endgroup$ – Adam Latosiński Apr 21 at 21:31
  • $\begingroup$ Totally right. I oversaw this missing factor. will adjust correspondingly. Thank you. $\endgroup$ – trancelocation Apr 22 at 6:47
  • 1
    $\begingroup$ @james2018 : I had to adjust the answer as the distribution in you question seems to miss a normalization factor depending on how one interprets it. If the distribution refers to $v$, then the distribution needs normalization. If it refers to $v/v_{avg}$, then all is good but integration limits need to be from 0 to 2. $\endgroup$ – trancelocation Apr 22 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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