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.

Re: Cauchy distribuation and maths help

id appreciate if someone could even point me in the right direction, im abit stumped...

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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

$\displaystyle y=I\tan(\theta)$

If we take the derivative we get

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

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

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

Now substitute tangent out for y to get

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

Solving for $\displaystyle d\theta$ gives

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

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.

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.

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 $\displaystyle \mu = 0 $

Then we need to calculate the integral

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

So now try to calculate this integral. What do you get?

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 ;)