# Cauchy distribuation and maths help

• Oct 9th 2012, 10:05 AM
bobby84
Cauchy distribuation and maths help
Hello i have this question that i am struggling to answer regarding changing variables, i really dont know how to do it :(

A gun fires at random in the angular range
−π/2 < θ < π/2 towards a wall a distance 'l' away. If 'y' is the coordinate along the wall,

show that

g(y)dy=(1/π) (1/(1 +(y/l)2)) (dy/l)

This is the Cauchy distribution. Assuming that the mean should be zero from symmetry considerations, try to find the standard deviation; what problem do you have? Truncate the distribution at a distance |y| = L either side of the peak. Calculate the new normalisation constant, and find the standard deviation.
• Oct 9th 2012, 01:10 PM
bobby84
Re: Cauchy distribuation and maths help
id appreciate if someone could even point me in the right direction, im abit stumped...
• Oct 9th 2012, 03:54 PM
TheEmptySet
Re: Cauchy distribuation and maths help
Quote:

Originally Posted by bobby84
Hello i have this question that i am struggling to answer regarding changing variables, i really dont know how to do it :(

A gun fires at random in the angular range
−π/2 < θ < π/2 towards a wall a distance 'l' away. If 'y' is the coordinate along the wall,

show that

g(y)dy=(1/π) (1/(1 +(y/l)2)) (dy/l)

This is the Cauchy distribution. Assuming that the mean should be zero from symmetry considerations, try to find the standard deviation; what problem do you have? Truncate the distribution at a distance |y| = L either side of the peak. Calculate the new normalisation constant, and find the standard deviation.

I am not 100% sure what you are asking, but I think this is what you mean....

Attachment 25139

From the diagram we see that

$y=I\tan(\theta)$

If we take the derivative we get

$dy=I\sec^2(\theta)d\theta$

We can rewrite the secant function interms of tangent using the pythagorean identity

$dy=I(1+\tan^2(\theta) d \theta$

Now substitute tangent out for y to get

$dy=I\left(1+\left(\frac{y}{I}\right)^2\right) d \theta$

Solving for $d\theta$ gives

$d\theta=\frac{1}{1+\left(\frac{y}{I}\right)^2} \frac{dy}{I}$

So for this to be a distribution the integral over the entire real line must be 1. You can use this to find the constant.
• Oct 9th 2012, 04:38 PM
bobby84
Re: Cauchy distribuation and maths help
im not sure if you read this bit...

This is the Cauchy distribution. Assuming that the mean should be zero from symmetry considerations, try to find the standard deviation; what problem do you have? Truncate the distribution at a distance |y| = L either side of the peak. Calculate the new normalisation constant, and find the standard deviation.
• Oct 9th 2012, 09:01 PM
TheEmptySet
Re: Cauchy distribuation and maths help
Quote:

Originally Posted by bobby84
im not sure if you read this bit...

This is the Cauchy distribution. Assuming that the mean should be zero from symmetry considerations, try to find the standard deviation; what problem do you have? Truncate the distribution at a distance |y| = L either side of the peak. Calculate the new normalisation constant, and find the standard deviation.

Well if $\mu = 0$

Then we need to calculate the integral

$\sigma^2=\frac{1}{\pi}\int_{-\infty}^{\infty}y^2\frac{1}{1+\left( \frac{y}{I}\right)^2}\frac{dy}{I}$

So now try to calculate this integral. What do you get?
• Oct 11th 2012, 03:17 AM
bobby84
Re: Cauchy distribuation and maths help
hey i got there in the end actually,

my final solution was....

(sigma) = sqrt(4-(pie)/2(pie)) l

cheers for the help ;)