Mixing
Functions of discrete random variables: inverting Poisson
$begingroup$
Let $X$ be a discrete random variable having the Poisson distribution
with parameter $lambda$ and let $Y=|sin (frac{pi}{2}X)|$. Find the
mass function of $Y$.
We are given the probability mass function of $X$:
$$
mathbb{P}(X=x) = frac{lambda^x}{x!}e^{-lambda}.
$$
It will be useful to express $X$ in terms of the new random variable $Y$:
$$
X = begin{cases}
frac{2}{pi} arcsin Y qquad & text{if } 0le Y le 1 \
0 qquad &text{otherwise}.
end{cases}
$$
We need to remove the negative part of the arcsine since $Y$ is defined as an absolute value.
Let me try to compute the probability mass function of $Y$:
$$
p_Y(y) = mathbb{P}(Y=y) = mathbb{P}(|sin (frac{pi}{2} X)|=y) =
begin{cases}
frac{lambda^{frac{2}{pi}arcsin y}}{(frac{2}{pi}arcsin y)!} e^{-lambda} & text{if } 0le y le 1\
e^{-lambda} & text{if } y < 0.
end{cases}
$$
I think my attempt is wrong because the probability mass function for $Y$ I just derived shows a factorial of a number that is highly unlikely to be a natural number.
Edit: after Frpzzd's answer, I guess the correct answer would be:
$$
p_Y(y) =
begin{cases}
lambda e^{-lambda} &text{if } y=sinfrac{pi}{2}=1 \
e^{-lambda} & text{if } y=0.
end{cases}
$$
I am still puzzled though, as $p_Y$ does not sum up to 1.
Edit 2: After Did's comment, I think I see what is going on now. Since $Xinmathbb{N}$, we see that $Y$ can only take the values $0$ or $1$, and I now edited my previous question to show this.
To conclude the solution, I guess that we just have to redefine the probability distribution as:
$$
p_Y(y) = begin{cases}
frac{lambda}{1+lambda} & text{if }y=1 \
frac{1}{1+lambda} & text{if }y=0,
end{cases}
$$
so that $p_Y$ does sum up to 1?
probability
asked Jan 17 at 22:48
J. D.J. D.
1878
$endgroup$
add a comment |
$begingroup$
Let $X$ be a discrete random variable having the Poisson distribution
with parameter $lambda$ and let $Y=|sin (frac{pi}{2}X)|$. Find the
mass function of $Y$.
We are given the probability mass function of $X$:
$$
mathbb{P}(X=x) = frac{lambda^x}{x!}e^{-lambda}.
$$
It will be useful to express $X$ in terms of the new random variable $Y$:
$$
X = begin{cases}
frac{2}{pi} arcsin Y qquad & text{if } 0le Y le 1 \
0 qquad &text{otherwise}.
end{cases}
$$
We need to remove the negative part of the arcsine since $Y$ is defined as an absolute value.
Let me try to compute the probability mass function of $Y$:
$$
p_Y(y) = mathbb{P}(Y=y) = mathbb{P}(|sin (frac{pi}{2} X)|=y) =
begin{cases}
frac{lambda^{frac{2}{pi}arcsin y}}{(frac{2}{pi}arcsin y)!} e^{-lambda} & text{if } 0le y le 1\
e^{-lambda} & text{if } y < 0.
end{cases}
$$
I think my attempt is wrong because the probability mass function for $Y$ I just derived shows a factorial of a number that is highly unlikely to be a natural number.
Edit: after Frpzzd's answer, I guess the correct answer would be:
$$
p_Y(y) =
begin{cases}
lambda e^{-lambda} &text{if } y=sinfrac{pi}{2}=1 \
e^{-lambda} & text{if } y=0.
end{cases}
$$
I am still puzzled though, as $p_Y$ does not sum up to 1.
Edit 2: After Did's comment, I think I see what is going on now. Since $Xinmathbb{N}$, we see that $Y$ can only take the values $0$ or $1$, and I now edited my previous question to show this.
To conclude the solution, I guess that we just have to redefine the probability distribution as:
$$
p_Y(y) = begin{cases}
frac{lambda}{1+lambda} & text{if }y=1 \
frac{1}{1+lambda} & text{if }y=0,
end{cases}
$$
so that $p_Y$ does sum up to 1?
probability
asked Jan 17 at 22:48
J. D.J. D.
1878
$endgroup$
$begingroup$
Hmmm... Even after having been pointed to the fact, you still believe that $P(Y=sin(2/pi))ne0$ where $Y=|sin(pi X/2)|$?
$endgroup$
– Did
Jan 18 at 10:58
add a comment |
$begingroup$
Let $X$ be a discrete random variable having the Poisson distribution
with parameter $lambda$ and let $Y=|sin (frac{pi}{2}X)|$. Find the
mass function of $Y$.
We are given the probability mass function of $X$:
$$
mathbb{P}(X=x) = frac{lambda^x}{x!}e^{-lambda}.
$$
It will be useful to express $X$ in terms of the new random variable $Y$:
$$
X = begin{cases}
frac{2}{pi} arcsin Y qquad & text{if } 0le Y le 1 \
0 qquad &text{otherwise}.
end{cases}
$$
We need to remove the negative part of the arcsine since $Y$ is defined as an absolute value.
Let me try to compute the probability mass function of $Y$:
$$
p_Y(y) = mathbb{P}(Y=y) = mathbb{P}(|sin (frac{pi}{2} X)|=y) =
begin{cases}
frac{lambda^{frac{2}{pi}arcsin y}}{(frac{2}{pi}arcsin y)!} e^{-lambda} & text{if } 0le y le 1\
e^{-lambda} & text{if } y < 0.
end{cases}
$$
I think my attempt is wrong because the probability mass function for $Y$ I just derived shows a factorial of a number that is highly unlikely to be a natural number.
Edit: after Frpzzd's answer, I guess the correct answer would be:
$$
p_Y(y) =
begin{cases}
lambda e^{-lambda} &text{if } y=sinfrac{pi}{2}=1 \
e^{-lambda} & text{if } y=0.
end{cases}
$$
I am still puzzled though, as $p_Y$ does not sum up to 1.
Edit 2: After Did's comment, I think I see what is going on now. Since $Xinmathbb{N}$, we see that $Y$ can only take the values $0$ or $1$, and I now edited my previous question to show this.
To conclude the solution, I guess that we just have to redefine the probability distribution as:
$$
p_Y(y) = begin{cases}
frac{lambda}{1+lambda} & text{if }y=1 \
frac{1}{1+lambda} & text{if }y=0,
end{cases}
$$
so that $p_Y$ does sum up to 1?
probability
asked Jan 17 at 22:48
J. D.J. D.
1878
$endgroup$
Let $X$ be a discrete random variable having the Poisson distribution
with parameter $lambda$ and let $Y=|sin (frac{pi}{2}X)|$. Find the
mass function of $Y$.
We are given the probability mass function of $X$:
$$
mathbb{P}(X=x) = frac{lambda^x}{x!}e^{-lambda}.
$$
It will be useful to express $X$ in terms of the new random variable $Y$:
$$
X = begin{cases}
frac{2}{pi} arcsin Y qquad & text{if } 0le Y le 1 \
0 qquad &text{otherwise}.
end{cases}
$$
We need to remove the negative part of the arcsine since $Y$ is defined as an absolute value.
Let me try to compute the probability mass function of $Y$:
$$
p_Y(y) = mathbb{P}(Y=y) = mathbb{P}(|sin (frac{pi}{2} X)|=y) =
begin{cases}
frac{lambda^{frac{2}{pi}arcsin y}}{(frac{2}{pi}arcsin y)!} e^{-lambda} & text{if } 0le y le 1\
e^{-lambda} & text{if } y < 0.
end{cases}
$$
I think my attempt is wrong because the probability mass function for $Y$ I just derived shows a factorial of a number that is highly unlikely to be a natural number.
Edit: after Frpzzd's answer, I guess the correct answer would be:
$$
p_Y(y) =
begin{cases}
lambda e^{-lambda} &text{if } y=sinfrac{pi}{2}=1 \
e^{-lambda} & text{if } y=0.
end{cases}
$$
I am still puzzled though, as $p_Y$ does not sum up to 1.
Edit 2: After Did's comment, I think I see what is going on now. Since $Xinmathbb{N}$, we see that $Y$ can only take the values $0$ or $1$, and I now edited my previous question to show this.
To conclude the solution, I guess that we just have to redefine the probability distribution as:
$$
p_Y(y) = begin{cases}
frac{lambda}{1+lambda} & text{if }y=1 \
frac{1}{1+lambda} & text{if }y=0,
end{cases}
$$
so that $p_Y$ does sum up to 1?
probability
probability
asked Jan 17 at 22:48
J. D.J. D.
1878
asked Jan 17 at 22:48
J. D.J. D.
1878
edited Jan 18 at 16:29
J. D.
asked Jan 17 at 22:48
J. D.J. D.
1878
asked Jan 17 at 22:48
J. D.J. D.
1878
asked Jan 17 at 22:48
J. D.J. D.
1878
1878
$begingroup$
Hmmm... Even after having been pointed to the fact, you still believe that $P(Y=sin(2/pi))ne0$ where $Y=|sin(pi X/2)|$?
$endgroup$
– Did
Jan 18 at 10:58
add a comment |
$begingroup$
Hmmm... Even after having been pointed to the fact, you still believe that $P(Y=sin(2/pi))ne0$ where $Y=|sin(pi X/2)|$?
$endgroup$
– Did
Jan 18 at 10:58
$begingroup$
Hmmm... Even after having been pointed to the fact, you still believe that $P(Y=sin(2/pi))ne0$ where $Y=|sin(pi X/2)|$?
$endgroup$
– Did
Jan 18 at 10:58
$begingroup$
Hmmm... Even after having been pointed to the fact, you still believe that $P(Y=sin(2/pi))ne0$ where $Y=|sin(pi X/2)|$?
$endgroup$
– Did
Jan 18 at 10:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your attempt is almost correct, but your PMF is not defined for all values of $y$ between $0$ and $1$; it is only defined for those values of $y$ for which $arcsin(y)inmathbb N$, since $Xinmathbb N$.Thus, you only need to tweak the conditions on your piecewise definition of $p_Y(y)$ a little bit.
answered Jan 17 at 22:56
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
I guess you meant "for those values of $y$ for which $frac{2}{pi}arcsin(y)inmathbb{N}$"?
$endgroup$
– J. D.
Jan 18 at 16:30
$begingroup$
@J.D. Right. $spacespace$
$endgroup$
– Frpzzd
Jan 18 at 23:02
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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$begingroup$
Your attempt is almost correct, but your PMF is not defined for all values of $y$ between $0$ and $1$; it is only defined for those values of $y$ for which $arcsin(y)inmathbb N$, since $Xinmathbb N$.Thus, you only need to tweak the conditions on your piecewise definition of $p_Y(y)$ a little bit.
answered Jan 17 at 22:56
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
I guess you meant "for those values of $y$ for which $frac{2}{pi}arcsin(y)inmathbb{N}$"?
$endgroup$
– J. D.
Jan 18 at 16:30
$begingroup$
@J.D. Right. $spacespace$
$endgroup$
– Frpzzd
Jan 18 at 23:02
add a comment |
$begingroup$
Your attempt is almost correct, but your PMF is not defined for all values of $y$ between $0$ and $1$; it is only defined for those values of $y$ for which $arcsin(y)inmathbb N$, since $Xinmathbb N$.Thus, you only need to tweak the conditions on your piecewise definition of $p_Y(y)$ a little bit.
answered Jan 17 at 22:56
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
I guess you meant "for those values of $y$ for which $frac{2}{pi}arcsin(y)inmathbb{N}$"?
$endgroup$
– J. D.
Jan 18 at 16:30
$begingroup$
@J.D. Right. $spacespace$
$endgroup$
– Frpzzd
Jan 18 at 23:02
add a comment |
$begingroup$
Your attempt is almost correct, but your PMF is not defined for all values of $y$ between $0$ and $1$; it is only defined for those values of $y$ for which $arcsin(y)inmathbb N$, since $Xinmathbb N$.Thus, you only need to tweak the conditions on your piecewise definition of $p_Y(y)$ a little bit.
answered Jan 17 at 22:56
FrpzzdFrpzzd
23k841109
$endgroup$
Your attempt is almost correct, but your PMF is not defined for all values of $y$ between $0$ and $1$; it is only defined for those values of $y$ for which $arcsin(y)inmathbb N$, since $Xinmathbb N$.Thus, you only need to tweak the conditions on your piecewise definition of $p_Y(y)$ a little bit.
answered Jan 17 at 22:56
FrpzzdFrpzzd
23k841109
answered Jan 17 at 22:56
FrpzzdFrpzzd
23k841109
answered Jan 17 at 22:56
FrpzzdFrpzzd
23k841109
answered Jan 17 at 22:56
FrpzzdFrpzzd
23k841109
23k841109
$begingroup$
I guess you meant "for those values of $y$ for which $frac{2}{pi}arcsin(y)inmathbb{N}$"?
$endgroup$
– J. D.
Jan 18 at 16:30
$begingroup$
@J.D. Right. $spacespace$
$endgroup$
– Frpzzd
Jan 18 at 23:02
add a comment |
$begingroup$
I guess you meant "for those values of $y$ for which $frac{2}{pi}arcsin(y)inmathbb{N}$"?
$endgroup$
– J. D.
Jan 18 at 16:30
$begingroup$
@J.D. Right. $spacespace$
$endgroup$
– Frpzzd
Jan 18 at 23:02
$begingroup$
I guess you meant "for those values of $y$ for which $frac{2}{pi}arcsin(y)inmathbb{N}$"?
$endgroup$
– J. D.
Jan 18 at 16:30
$begingroup$
I guess you meant "for those values of $y$ for which $frac{2}{pi}arcsin(y)inmathbb{N}$"?
$endgroup$
– J. D.
Jan 18 at 16:30
$begingroup$
@J.D. Right. $spacespace$
$endgroup$
– Frpzzd
Jan 18 at 23:02
$begingroup$
@J.D. Right. $spacespace$
$endgroup$
– Frpzzd
Jan 18 at 23:02
add a comment |
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$begingroup$
Hmmm... Even after having been pointed to the fact, you still believe that $P(Y=sin(2/pi))ne0$ where $Y=|sin(pi X/2)|$?
$endgroup$
– Did
Jan 18 at 10:58