Finding the PMF of Y with Known X

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SUMMARY

The discussion focuses on finding the Probability Mass Function (PMF) of the random variable Y, defined as \(Y = \sin\Big(\frac{\pi}{2X}\Big)\), given the PMF of X, \(p_X[k] = \frac{1}{5}\) for \(k = 0, 1, 2, 3, 4\). Participants highlight the transformation of X into Y and seek clarification on the interpretation of the PMF notation. The problem is noted to be previously solved, yet the specific solution remains unprovided in the discussion.

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  • Understanding of Probability Mass Functions (PMFs)
  • Knowledge of transformations of random variables
  • Familiarity with trigonometric functions and their properties
  • Basic concepts of characteristic functions in probability theory
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Dustinsfl
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How do I find the PMF of Y when I know X?
\[
Y = \sin\Big(\frac{\pi}{2X}\Big)
\]
and
\[
p_X[k] = \frac{1}{5}
\]
for \(k = 0,1,\ldots, 4\).
 
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Maybe there're several ways to solve this problem. It's clear that $Y$ is a transformation of $X$. So have you seen any theorems about characteristic funtions of transformated random variables?

EDIT:
I now see that this problem is already solved.
 
Siron said:
...EDIT:
I now see that this problem is already solved.

Well, the [SOLVED] prefix has been added, but I see no solution...
 
dwsmith said:
How do I find the PMF of Y when I know X?
\[
Y = \sin\Big(\frac{\pi}{2X}\Big)
\]
and
\[
p_X[k] = \frac{1}{5}
\]
for \(k = 0,1,\ldots, 4\).

There is a point not clear: writing $\displaystyle p_X[k] = \frac{1}{5}, k= 0,1,2,3,4$ means writing $\displaystyle P\{X=k \} = \frac{1}{5}, k= 0,1,2,3,4$?...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
There is a point not clear: writing $\displaystyle p_X[k] = \frac{1}{5}, k= 0,1,2,3,4$ means writing $\displaystyle P\{X=k \} = \frac{1}{5}, k= 0,1,2,3,4$?...

Kind regards

$\chi$ $\sigma$

The reason of my question is that for X=0 defining $\displaystyle Y = \sin \frac{\pi}{2\ X}$ is a little difficult task (Dull)...

Kind regards

$\chi$ $\sigma$
 

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