MHB Finding the PMF of Y with Known X

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To find the PMF of Y given X, where Y is defined as Y = sin(π/(2X)) and p_X[k] = 1/5 for k = 0, 1, 2, 3, 4, one must consider the transformation of the random variable X. The discussion highlights the challenge of defining Y when X equals zero, as it leads to an undefined expression. Participants express confusion regarding the notation p_X[k] and its implication for the probability distribution of X. The thread indicates that the problem has been previously addressed, yet no clear solution is provided within the discussion. Overall, the focus remains on understanding the transformation and its implications for calculating the PMF of Y.
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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