MHB What are the distributions for R and fi in polar coordinates?

[Nina22]

Hello, I have a problem. I have two independent variables, X and Y. I use polar coordinates so X=r*cos(fi) and Y=r*sin(fi)

R >= 0 and -pi <= fi <= pi

I got two distributions: for R is: r*e^((-r^2)/2)
and for fi: 1/2pi

but now I don't know which kind of distribution is for each of them (I don't know the names of the distributions)

Please help me

 

[I like Serena]

Hello, I have a problem. I have two independent variables, X and Y. I use polar coordinates so X=r*cos(fi) and Y=r*sin(fi)

R >= 0 and -pi <= fi <= pi

I got two distributions: for R is: r*e^((-r^2)/2)
and for fi: 1/2pi

but now I don't know which kind of distribution is for each of them (I don't know the names of the distributions)

Please help me

Welcome to MHB Nina! :)The relevant distributions are:

• Uniform distribution, identified with a lower bound $a$ and an upper bound $b$.
• Normal distribution, identified with a mean $\mu$ and a standard deviation $\sigma$.
• Weibull distribution, with parameters $k$ and $\lambda$.

Can you match them with your random variables and find their identifying parameters?
Or otherwise show some of your thoughts?

 

[chisigma]

Hello, I have a problem. I have two independent variables, X and Y. I use polar coordinates so X=r*cos(fi) and Y=r*sin(fi)

R >= 0 and -pi <= fi <= pi

I got two distributions: for R is: r*e^((-r^2)/2)
and for fi: 1/2pi

but now I don't know which kind of distribution is for each of them (I don't know the names of the distributions)

Please help me

Wellcome to a new Italian member on MHB!... congratulation to Nina!...

If You consider to have a vector V(X,Y) where X and Y and normal random variables with zero mean a variance $\displaystyle \sigma^{2}$, then both X and Y have p.d.f. ... $\displaystyle f(u) = \frac{1}{\sqrt{2\ \pi}\ \sigma}\ e^{- \frac{u^{2}}{2\ \sigma^{2}}}\ (1)$

In such a case You can demonstrate that the angle of V is uniformly distributed between $- \pi$ and $\pi$ and the module |V| of the vector is Rayleigh distributed, i.e. |V| fas p.d.f. ...$\displaystyle g(w) = \frac{w}{\sigma^{2}}\ e^ {- \frac{w^{2}}{2\ \sigma^{2}}}\ (2)$

In Your case X and Y are normal distributed r.v. with zero mean and variance $\displaystyle \sigma^{2}= 1$...

Kind regards

$\chi$ $\sigma$

 

[Nina22]

Thank you very much for your answers!

I like Serena: I was thinking and didn't know if it was right that fi is distributed uniformely

 

[I like Serena]

Thank you very much for your answers!

I like Serena: I was thinking and didn't know if it was right that fi is distributed uniformely

Yes. That is correct!

Its distribution function is $\frac{1}{2\pi}$, which is constant.
That is the prime characteristic of a uniform distribution function: whatever the outcome, the probability is the same.
Can you tell what its lower and upper bounds are?
With the right bounds the total cumulative probability has to come out as $1$.

 

[Nina22]

So I was right :) Thank you

and R has the Rayleigh distribution

 

[I like Serena]

So I was right :) Thank you

and R has the Rayleigh distribution

Good! :D