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Asked Oct 12, 2019
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QUESTION 1
1 +x
x
1. Find the pdf of the random vector (R,T
a) Suppose (X, Y) has the pdf f (x, y) =
Rcos (T) Y Rsin(T)
which is (X, Y) expressed in polar coordinates, that is, X =
1
Let X follow the Cauchy distribution with the pdf fx(x)
1
b)
Find the pdf of 1/X
E R.
12
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QUESTION 1 1 +x x 1. Find the pdf of the random vector (R,T a) Suppose (X, Y) has the pdf f (x, y) = Rcos (T) Y Rsin(T) which is (X, Y) expressed in polar coordinates, that is, X = 1 Let X follow the Cauchy distribution with the pdf fx(x) 1 b) Find the pdf of 1/X E R. 12

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Expert Answer

Step 1

Note:

Hi! Thanks for posting the question. Although the question is presented as one question with two parts, each part is quite lengthy and can be treated as a separate question. Also, solution to any one part is independent of that of the other parts. Hence, we have treated this as two separate questions and according to our policy, have solved the first question for you. If you need help with any of the other parts, please re-post the question and mention the part you need help with.

a)

The joint pdf of (x,y) is given by –

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1+ x ,where x2 y2 <1 f (x, y) TT 0, Otherwise

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Step 2

Finding the p.d.f of the random vector (R, T)

Given That X and Y in the form of polar coordinates

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R cos(T) X ох дх -R sin(T) cos(T), дт R sin(T) Y дү дY sin (T) R cos(T) дт Now finding the value of T R sin(T) Y tan T R cos(T) X => T = tan -1 where 0 T 2t Now finding the value of R x22 R2 cos2 (T) R?sin2(T) = R2(cos2 (T) sin2 (T)) = R2 x 1 Because cos2 (T)+ sin2(T) = 1 X2Y2 R

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Step 3

Now we have to convert the X and Y in the form of ...

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гәх әх DR дт дү дү LaR aT Ә(х, Ү) a (R,T) [cos (T) sin(T) -R sin (T] R cos(T) > ] = R(cos2 (T) sin2(T)) = R Now, the joint pdf of (R, T) is given by g(R,T) f(R cos(T), R sin(T) x \J] (1 R cos(T)) x R => g(R,T) п The p.d.f of random vector g(R,T) is (1R cos(T))R 0 R 1 and 0 < T< Z g(R,T) п otherwise 0;

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