Let $U sim textrm{Unif}(0, pi/ 2)$. Find the PDF of $sin(U)$.
$begingroup$
This is almost the same as Suppose that X ∼ U ( $− π/2$ , $π/2$ ) . Find the pdf of Y = tan(X)., but making sure I am understanding the process:
Let $U sim textrm{Unif}(0, pi/ 2)$. Find the PDF of $sin(U)$.
$$begin{align}
F_Y(y) &= P(Y<y) = P(sin(u)<y) = P(u < sin^{-1}(y)) \
&=F_U(sin^{-1}(y)) = int_0^{sin^{-1}(y)}frac{1}{frac{pi}{2}}du \
&= frac{1}{frac{pi}{2}}u |_0^{sin^{-1}(y)} \
&= frac{2}{pi}sin^{-1}(y) \
end{align} $$
That is the CDF. To find the PDF, take the derivative with respect to $y$ to get:
$$ frac{2}{pi}frac{1}{sqrt{1-y^2}} $$
- Is the work & logic correct?
- Is the support of $Y$ the same as the support for U: $[0,pi/2]$?
random-variables uniform-distribution
edited Dec 8 '18 at 22:11
Andrews
4031317
asked Dec 8 '18 at 21:56
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$endgroup$
add a comment |
$begingroup$
This is almost the same as Suppose that X ∼ U ( $− π/2$ , $π/2$ ) . Find the pdf of Y = tan(X)., but making sure I am understanding the process:
Let $U sim textrm{Unif}(0, pi/ 2)$. Find the PDF of $sin(U)$.
$$begin{align}
F_Y(y) &= P(Y<y) = P(sin(u)<y) = P(u < sin^{-1}(y)) \
&=F_U(sin^{-1}(y)) = int_0^{sin^{-1}(y)}frac{1}{frac{pi}{2}}du \
&= frac{1}{frac{pi}{2}}u |_0^{sin^{-1}(y)} \
&= frac{2}{pi}sin^{-1}(y) \
end{align} $$
That is the CDF. To find the PDF, take the derivative with respect to $y$ to get:
$$ frac{2}{pi}frac{1}{sqrt{1-y^2}} $$
- Is the work & logic correct?
- Is the support of $Y$ the same as the support for U: $[0,pi/2]$?
random-variables uniform-distribution
edited Dec 8 '18 at 22:11
Andrews
4031317
asked Dec 8 '18 at 21:56
user603569user603569
728
$endgroup$
2
$begingroup$
The support for $Y$ is $[0,1]$. Always good to confirm that the PDF integrates to 1 over $[0,1]$. In this case it does. You can compute the integral by substituting $y = sin theta$.
$endgroup$
– Aditya Dua
Dec 8 '18 at 22:21
add a comment |
$begingroup$
This is almost the same as Suppose that X ∼ U ( $− π/2$ , $π/2$ ) . Find the pdf of Y = tan(X)., but making sure I am understanding the process:
Let $U sim textrm{Unif}(0, pi/ 2)$. Find the PDF of $sin(U)$.
$$begin{align}
F_Y(y) &= P(Y<y) = P(sin(u)<y) = P(u < sin^{-1}(y)) \
&=F_U(sin^{-1}(y)) = int_0^{sin^{-1}(y)}frac{1}{frac{pi}{2}}du \
&= frac{1}{frac{pi}{2}}u |_0^{sin^{-1}(y)} \
&= frac{2}{pi}sin^{-1}(y) \
end{align} $$
That is the CDF. To find the PDF, take the derivative with respect to $y$ to get:
$$ frac{2}{pi}frac{1}{sqrt{1-y^2}} $$
- Is the work & logic correct?
- Is the support of $Y$ the same as the support for U: $[0,pi/2]$?
random-variables uniform-distribution
edited Dec 8 '18 at 22:11
Andrews
4031317
asked Dec 8 '18 at 21:56
user603569user603569
728
$endgroup$
This is almost the same as Suppose that X ∼ U ( $− π/2$ , $π/2$ ) . Find the pdf of Y = tan(X)., but making sure I am understanding the process:
Let $U sim textrm{Unif}(0, pi/ 2)$. Find the PDF of $sin(U)$.
$$begin{align}
F_Y(y) &= P(Y<y) = P(sin(u)<y) = P(u < sin^{-1}(y)) \
&=F_U(sin^{-1}(y)) = int_0^{sin^{-1}(y)}frac{1}{frac{pi}{2}}du \
&= frac{1}{frac{pi}{2}}u |_0^{sin^{-1}(y)} \
&= frac{2}{pi}sin^{-1}(y) \
end{align} $$
That is the CDF. To find the PDF, take the derivative with respect to $y$ to get:
$$ frac{2}{pi}frac{1}{sqrt{1-y^2}} $$
- Is the work & logic correct?
- Is the support of $Y$ the same as the support for U: $[0,pi/2]$?
random-variables uniform-distribution
random-variables uniform-distribution
edited Dec 8 '18 at 22:11
Andrews
4031317
asked Dec 8 '18 at 21:56
user603569user603569
728
edited Dec 8 '18 at 22:11
Andrews
4031317
asked Dec 8 '18 at 21:56
user603569user603569
728
edited Dec 8 '18 at 22:11
Andrews
4031317
edited Dec 8 '18 at 22:11
Andrews
4031317
edited Dec 8 '18 at 22:11
Andrews
4031317
4031317
asked Dec 8 '18 at 21:56
user603569user603569
728
asked Dec 8 '18 at 21:56
user603569user603569
728
asked Dec 8 '18 at 21:56
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2
$begingroup$
The support for $Y$ is $[0,1]$. Always good to confirm that the PDF integrates to 1 over $[0,1]$. In this case it does. You can compute the integral by substituting $y = sin theta$.
$endgroup$
– Aditya Dua
Dec 8 '18 at 22:21
add a comment |
2
$begingroup$
The support for $Y$ is $[0,1]$. Always good to confirm that the PDF integrates to 1 over $[0,1]$. In this case it does. You can compute the integral by substituting $y = sin theta$.
$endgroup$
– Aditya Dua
Dec 8 '18 at 22:21
2
2
$begingroup$
The support for $Y$ is $[0,1]$. Always good to confirm that the PDF integrates to 1 over $[0,1]$. In this case it does. You can compute the integral by substituting $y = sin theta$.
$endgroup$
– Aditya Dua
Dec 8 '18 at 22:21
$begingroup$
The support for $Y$ is $[0,1]$. Always good to confirm that the PDF integrates to 1 over $[0,1]$. In this case it does. You can compute the integral by substituting $y = sin theta$.
$endgroup$
– Aditya Dua
Dec 8 '18 at 22:21
add a comment |
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$begingroup$
Per Aditya Dua: The support for Y is [0,1]. Always good to confirm that the PDF integrates to 1 over [0,1]. In this case it does. You can compute the integral by substituting y=sinθ.
answered Dec 8 '18 at 23:25
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$begingroup$
Per Aditya Dua: The support for Y is [0,1]. Always good to confirm that the PDF integrates to 1 over [0,1]. In this case it does. You can compute the integral by substituting y=sinθ.
answered Dec 8 '18 at 23:25
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Per Aditya Dua: The support for Y is [0,1]. Always good to confirm that the PDF integrates to 1 over [0,1]. In this case it does. You can compute the integral by substituting y=sinθ.
answered Dec 8 '18 at 23:25
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$begingroup$
Per Aditya Dua: The support for Y is [0,1]. Always good to confirm that the PDF integrates to 1 over [0,1]. In this case it does. You can compute the integral by substituting y=sinθ.
answered Dec 8 '18 at 23:25
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Per Aditya Dua: The support for Y is [0,1]. Always good to confirm that the PDF integrates to 1 over [0,1]. In this case it does. You can compute the integral by substituting y=sinθ.
answered Dec 8 '18 at 23:25
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answered Dec 8 '18 at 23:25
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answered Dec 8 '18 at 23:25
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The support for $Y$ is $[0,1]$. Always good to confirm that the PDF integrates to 1 over $[0,1]$. In this case it does. You can compute the integral by substituting $y = sin theta$.
$endgroup$
– Aditya Dua
Dec 8 '18 at 22:21