Mixing
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A segment is removed from a semi circle what is the pdf of the remaining area?
Clash Royale CLAN TAG#URR8PPP
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1
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Question:
Given the illustration below, the cdf of Z is
beginequation
Pr(Zleq z) = fractextexp(-lambda A_z)-textexp(-N/2)1-textexp(-N/2)
endequation
How could one Prove that
beginequation
f_z(Z) = frac2lambda sqrtR^2-z^21-textexp(-N/2) textexp[-lambda A_z]
endequation
for $0<zleq R$
where $A_z$ = $R^2 [cos^-1 (z/R)- (z/R)sqrt1-(z/R)^2]$
Challenge:
The first problem I have is understanding how $A_z$ was derived, then, how the numerator of the first equation of the pdf was gotten. The closest clue I have to $A_z$ is this question and this one, but here, the height of the segment here is unknown.
In case it might interest someone, $N = lambda pi R^2$ and $lambda$ represents the density of a two dimensional Poisson Point Process
probability probability-theory probability-distributions
asked Jul 24 at 6:50
Abdulhameed
54111
add a comment |Â
up vote
1
down vote
favorite
Question:
Given the illustration below, the cdf of Z is
beginequation
Pr(Zleq z) = fractextexp(-lambda A_z)-textexp(-N/2)1-textexp(-N/2)
endequation
How could one Prove that
beginequation
f_z(Z) = frac2lambda sqrtR^2-z^21-textexp(-N/2) textexp[-lambda A_z]
endequation
for $0<zleq R$
where $A_z$ = $R^2 [cos^-1 (z/R)- (z/R)sqrt1-(z/R)^2]$
Challenge:
The first problem I have is understanding how $A_z$ was derived, then, how the numerator of the first equation of the pdf was gotten. The closest clue I have to $A_z$ is this question and this one, but here, the height of the segment here is unknown.
In case it might interest someone, $N = lambda pi R^2$ and $lambda$ represents the density of a two dimensional Poisson Point Process
probability probability-theory probability-distributions
asked Jul 24 at 6:50
Abdulhameed
54111
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Question:
Given the illustration below, the cdf of Z is
beginequation
Pr(Zleq z) = fractextexp(-lambda A_z)-textexp(-N/2)1-textexp(-N/2)
endequation
How could one Prove that
beginequation
f_z(Z) = frac2lambda sqrtR^2-z^21-textexp(-N/2) textexp[-lambda A_z]
endequation
for $0<zleq R$
where $A_z$ = $R^2 [cos^-1 (z/R)- (z/R)sqrt1-(z/R)^2]$
Challenge:
The first problem I have is understanding how $A_z$ was derived, then, how the numerator of the first equation of the pdf was gotten. The closest clue I have to $A_z$ is this question and this one, but here, the height of the segment here is unknown.
In case it might interest someone, $N = lambda pi R^2$ and $lambda$ represents the density of a two dimensional Poisson Point Process
probability probability-theory probability-distributions
asked Jul 24 at 6:50
Abdulhameed
54111
Question:
Given the illustration below, the cdf of Z is
beginequation
Pr(Zleq z) = fractextexp(-lambda A_z)-textexp(-N/2)1-textexp(-N/2)
endequation
How could one Prove that
beginequation
f_z(Z) = frac2lambda sqrtR^2-z^21-textexp(-N/2) textexp[-lambda A_z]
endequation
for $0<zleq R$
where $A_z$ = $R^2 [cos^-1 (z/R)- (z/R)sqrt1-(z/R)^2]$
Challenge:
The first problem I have is understanding how $A_z$ was derived, then, how the numerator of the first equation of the pdf was gotten. The closest clue I have to $A_z$ is this question and this one, but here, the height of the segment here is unknown.
In case it might interest someone, $N = lambda pi R^2$ and $lambda$ represents the density of a two dimensional Poisson Point Process
probability probability-theory probability-distributions
asked Jul 24 at 6:50
Abdulhameed
54111
edited Jul 24 at 9:30
asked Jul 24 at 6:50
Abdulhameed
54111
asked Jul 24 at 6:50
Abdulhameed
54111
asked Jul 24 at 6:50
Abdulhameed
54111
54111
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
I understand that you have already recognised that this problem can be solved by using the identity relating the PDF of a distribution to the CDF: $ f_Z(z) = F_Z'(z)$,
which for your particular CDF is equivalent to:
$$ f_Z(z) = frac-lambda A'(z) exp(-lambda A(z))1- exp(-N/2).$$
So it remains to calculate $A(z)$, and then use this to derive $A'(z)$.
Calculating A(z)
In the following I refer to my diagram (unfortunately not as well produced as your graphic, but you should get the point).
Let $theta_z$ denote the angle made between $A hat O B$, using the standard trigonometric identify for the cosine we have that
$$theta_z = cos^-1(z/R).$$
Therefore the area of the circular wedge with angle $2theta_z$ (i.e. the wedge bounded by the lines $OB$ and $OC$) is
beginalignedW_z &= pi R^2 times frac2theta_z2 pi \
& = R^2 theta_z \
& = R^2 cos^-1(z/R).
endaligned
We can now obtain $A(z)$ by subtracting from $W_z$ the area, $T_z$, of the triangle made out by the lines $OBCO$, which is equal to the area of the rectangle with side lengths equal to those of the line segments $OA$ and $AB$.
Since $A$ has coordinates $(z,0)$ the line segment $OA$ has length $z$; we can calculate the coordinates of $B$ by relying on the formula $x^2 + y^2 = R^2$, which gives coordinates $(z, sqrtR^2 - z^2)$, and so the length of $AB$ is $sqrtR^2 - z^2$. Combining these we get
$$T_z = z sqrtR^2 - z^2$$,
and further more
beginaligned
A(z) &= W_z - T_z \
& = R^2 cos^-1(z/R) - z sqrtR^2 - z^2 \
& = R^2 left( cos^-1(z/R) - (z/R) sqrt1 - (z/R)^2 right),
endaligned
which is as you gave in the question statement.
Calculating $A'(z)$
This now follows from appealing to standard identities for derivatives. If we denote
$$g(x) = cos^-1(x) - x sqrt1 - x^2,$$
then we have $A(z) = R^2 g(z/R)$, and so using the chain rule:
beginaligned
A'(z) & = R^2 fracddz g(z/R) \
& = R^2 g'(z/R) fracddz left(fraczRright) \
& = R g'(z/R).
endaligned
So concentrating on $g(x)$ we differentiate this using the identity for the derivative of $cos^-1$, as well as the product rule to handle the second term to obtain:
$$g'(x) = - frac1sqrt1-x^2 - frac1-2x^2sqrt1-x^2, $$
where the first term above is contributed from $cos^-1$, and the second is from the product term. Rearranging this we have
beginaligned
g'(x) &= -2frac1 - x^2sqrt1-x^2 \
& = -2 sqrt1 - x^2.
endaligned
And so we have
beginaligned
A'(z) &= Rg'(z/R) \
& = -2R sqrt1 - (z/R)^2 \
& = -2 sqrtR^2 - z^2.
endaligned
Substituting this into the original formula provided for $f_Z(z)$ gives the desired answer.
answered Jul 24 at 18:27
owen88
3,2731021
Thanks for this brilliant answer. Please could you add some detail on how $A'(z) = R g'(z/R).$ was gotten using the chain rule. I didn't get that part well even with my basic knowledge of the chain rule.
– Abdulhameed
Jul 25 at 0:09
1
I have added a few lines breaking down the chain rule calculation.
– owen88
Jul 25 at 14:50
Those lines are well appreciated and understood. Thank you very much
– Abdulhameed
Jul 26 at 3:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I understand that you have already recognised that this problem can be solved by using the identity relating the PDF of a distribution to the CDF: $ f_Z(z) = F_Z'(z)$,
which for your particular CDF is equivalent to:
$$ f_Z(z) = frac-lambda A'(z) exp(-lambda A(z))1- exp(-N/2).$$
So it remains to calculate $A(z)$, and then use this to derive $A'(z)$.
Calculating A(z)
In the following I refer to my diagram (unfortunately not as well produced as your graphic, but you should get the point).
Let $theta_z$ denote the angle made between $A hat O B$, using the standard trigonometric identify for the cosine we have that
$$theta_z = cos^-1(z/R).$$
Therefore the area of the circular wedge with angle $2theta_z$ (i.e. the wedge bounded by the lines $OB$ and $OC$) is
beginalignedW_z &= pi R^2 times frac2theta_z2 pi \
& = R^2 theta_z \
& = R^2 cos^-1(z/R).
endaligned
We can now obtain $A(z)$ by subtracting from $W_z$ the area, $T_z$, of the triangle made out by the lines $OBCO$, which is equal to the area of the rectangle with side lengths equal to those of the line segments $OA$ and $AB$.
Since $A$ has coordinates $(z,0)$ the line segment $OA$ has length $z$; we can calculate the coordinates of $B$ by relying on the formula $x^2 + y^2 = R^2$, which gives coordinates $(z, sqrtR^2 - z^2)$, and so the length of $AB$ is $sqrtR^2 - z^2$. Combining these we get
$$T_z = z sqrtR^2 - z^2$$,
and further more
beginaligned
A(z) &= W_z - T_z \
& = R^2 cos^-1(z/R) - z sqrtR^2 - z^2 \
& = R^2 left( cos^-1(z/R) - (z/R) sqrt1 - (z/R)^2 right),
endaligned
which is as you gave in the question statement.
Calculating $A'(z)$
This now follows from appealing to standard identities for derivatives. If we denote
$$g(x) = cos^-1(x) - x sqrt1 - x^2,$$
then we have $A(z) = R^2 g(z/R)$, and so using the chain rule:
beginaligned
A'(z) & = R^2 fracddz g(z/R) \
& = R^2 g'(z/R) fracddz left(fraczRright) \
& = R g'(z/R).
endaligned
So concentrating on $g(x)$ we differentiate this using the identity for the derivative of $cos^-1$, as well as the product rule to handle the second term to obtain:
$$g'(x) = - frac1sqrt1-x^2 - frac1-2x^2sqrt1-x^2, $$
where the first term above is contributed from $cos^-1$, and the second is from the product term. Rearranging this we have
beginaligned
g'(x) &= -2frac1 - x^2sqrt1-x^2 \
& = -2 sqrt1 - x^2.
endaligned
And so we have
beginaligned
A'(z) &= Rg'(z/R) \
& = -2R sqrt1 - (z/R)^2 \
& = -2 sqrtR^2 - z^2.
endaligned
Substituting this into the original formula provided for $f_Z(z)$ gives the desired answer.
answered Jul 24 at 18:27
owen88
3,2731021
Thanks for this brilliant answer. Please could you add some detail on how $A'(z) = R g'(z/R).$ was gotten using the chain rule. I didn't get that part well even with my basic knowledge of the chain rule.
– Abdulhameed
Jul 25 at 0:09
1
I have added a few lines breaking down the chain rule calculation.
– owen88
Jul 25 at 14:50
Those lines are well appreciated and understood. Thank you very much
– Abdulhameed
Jul 26 at 3:49
add a comment |Â
up vote
1
down vote
accepted
I understand that you have already recognised that this problem can be solved by using the identity relating the PDF of a distribution to the CDF: $ f_Z(z) = F_Z'(z)$,
which for your particular CDF is equivalent to:
$$ f_Z(z) = frac-lambda A'(z) exp(-lambda A(z))1- exp(-N/2).$$
So it remains to calculate $A(z)$, and then use this to derive $A'(z)$.
Calculating A(z)
In the following I refer to my diagram (unfortunately not as well produced as your graphic, but you should get the point).
Let $theta_z$ denote the angle made between $A hat O B$, using the standard trigonometric identify for the cosine we have that
$$theta_z = cos^-1(z/R).$$
Therefore the area of the circular wedge with angle $2theta_z$ (i.e. the wedge bounded by the lines $OB$ and $OC$) is
beginalignedW_z &= pi R^2 times frac2theta_z2 pi \
& = R^2 theta_z \
& = R^2 cos^-1(z/R).
endaligned
We can now obtain $A(z)$ by subtracting from $W_z$ the area, $T_z$, of the triangle made out by the lines $OBCO$, which is equal to the area of the rectangle with side lengths equal to those of the line segments $OA$ and $AB$.
Since $A$ has coordinates $(z,0)$ the line segment $OA$ has length $z$; we can calculate the coordinates of $B$ by relying on the formula $x^2 + y^2 = R^2$, which gives coordinates $(z, sqrtR^2 - z^2)$, and so the length of $AB$ is $sqrtR^2 - z^2$. Combining these we get
$$T_z = z sqrtR^2 - z^2$$,
and further more
beginaligned
A(z) &= W_z - T_z \
& = R^2 cos^-1(z/R) - z sqrtR^2 - z^2 \
& = R^2 left( cos^-1(z/R) - (z/R) sqrt1 - (z/R)^2 right),
endaligned
which is as you gave in the question statement.
Calculating $A'(z)$
This now follows from appealing to standard identities for derivatives. If we denote
$$g(x) = cos^-1(x) - x sqrt1 - x^2,$$
then we have $A(z) = R^2 g(z/R)$, and so using the chain rule:
beginaligned
A'(z) & = R^2 fracddz g(z/R) \
& = R^2 g'(z/R) fracddz left(fraczRright) \
& = R g'(z/R).
endaligned
So concentrating on $g(x)$ we differentiate this using the identity for the derivative of $cos^-1$, as well as the product rule to handle the second term to obtain:
$$g'(x) = - frac1sqrt1-x^2 - frac1-2x^2sqrt1-x^2, $$
where the first term above is contributed from $cos^-1$, and the second is from the product term. Rearranging this we have
beginaligned
g'(x) &= -2frac1 - x^2sqrt1-x^2 \
& = -2 sqrt1 - x^2.
endaligned
And so we have
beginaligned
A'(z) &= Rg'(z/R) \
& = -2R sqrt1 - (z/R)^2 \
& = -2 sqrtR^2 - z^2.
endaligned
Substituting this into the original formula provided for $f_Z(z)$ gives the desired answer.
answered Jul 24 at 18:27
owen88
3,2731021
Thanks for this brilliant answer. Please could you add some detail on how $A'(z) = R g'(z/R).$ was gotten using the chain rule. I didn't get that part well even with my basic knowledge of the chain rule.
– Abdulhameed
Jul 25 at 0:09
1
I have added a few lines breaking down the chain rule calculation.
– owen88
Jul 25 at 14:50
Those lines are well appreciated and understood. Thank you very much
– Abdulhameed
Jul 26 at 3:49
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I understand that you have already recognised that this problem can be solved by using the identity relating the PDF of a distribution to the CDF: $ f_Z(z) = F_Z'(z)$,
which for your particular CDF is equivalent to:
$$ f_Z(z) = frac-lambda A'(z) exp(-lambda A(z))1- exp(-N/2).$$
So it remains to calculate $A(z)$, and then use this to derive $A'(z)$.
Calculating A(z)
In the following I refer to my diagram (unfortunately not as well produced as your graphic, but you should get the point).
Let $theta_z$ denote the angle made between $A hat O B$, using the standard trigonometric identify for the cosine we have that
$$theta_z = cos^-1(z/R).$$
Therefore the area of the circular wedge with angle $2theta_z$ (i.e. the wedge bounded by the lines $OB$ and $OC$) is
beginalignedW_z &= pi R^2 times frac2theta_z2 pi \
& = R^2 theta_z \
& = R^2 cos^-1(z/R).
endaligned
We can now obtain $A(z)$ by subtracting from $W_z$ the area, $T_z$, of the triangle made out by the lines $OBCO$, which is equal to the area of the rectangle with side lengths equal to those of the line segments $OA$ and $AB$.
Since $A$ has coordinates $(z,0)$ the line segment $OA$ has length $z$; we can calculate the coordinates of $B$ by relying on the formula $x^2 + y^2 = R^2$, which gives coordinates $(z, sqrtR^2 - z^2)$, and so the length of $AB$ is $sqrtR^2 - z^2$. Combining these we get
$$T_z = z sqrtR^2 - z^2$$,
and further more
beginaligned
A(z) &= W_z - T_z \
& = R^2 cos^-1(z/R) - z sqrtR^2 - z^2 \
& = R^2 left( cos^-1(z/R) - (z/R) sqrt1 - (z/R)^2 right),
endaligned
which is as you gave in the question statement.
Calculating $A'(z)$
This now follows from appealing to standard identities for derivatives. If we denote
$$g(x) = cos^-1(x) - x sqrt1 - x^2,$$
then we have $A(z) = R^2 g(z/R)$, and so using the chain rule:
beginaligned
A'(z) & = R^2 fracddz g(z/R) \
& = R^2 g'(z/R) fracddz left(fraczRright) \
& = R g'(z/R).
endaligned
So concentrating on $g(x)$ we differentiate this using the identity for the derivative of $cos^-1$, as well as the product rule to handle the second term to obtain:
$$g'(x) = - frac1sqrt1-x^2 - frac1-2x^2sqrt1-x^2, $$
where the first term above is contributed from $cos^-1$, and the second is from the product term. Rearranging this we have
beginaligned
g'(x) &= -2frac1 - x^2sqrt1-x^2 \
& = -2 sqrt1 - x^2.
endaligned
And so we have
beginaligned
A'(z) &= Rg'(z/R) \
& = -2R sqrt1 - (z/R)^2 \
& = -2 sqrtR^2 - z^2.
endaligned
Substituting this into the original formula provided for $f_Z(z)$ gives the desired answer.
answered Jul 24 at 18:27
owen88
3,2731021
I understand that you have already recognised that this problem can be solved by using the identity relating the PDF of a distribution to the CDF: $ f_Z(z) = F_Z'(z)$,
which for your particular CDF is equivalent to:
$$ f_Z(z) = frac-lambda A'(z) exp(-lambda A(z))1- exp(-N/2).$$
So it remains to calculate $A(z)$, and then use this to derive $A'(z)$.
Calculating A(z)
In the following I refer to my diagram (unfortunately not as well produced as your graphic, but you should get the point).
Let $theta_z$ denote the angle made between $A hat O B$, using the standard trigonometric identify for the cosine we have that
$$theta_z = cos^-1(z/R).$$
Therefore the area of the circular wedge with angle $2theta_z$ (i.e. the wedge bounded by the lines $OB$ and $OC$) is
beginalignedW_z &= pi R^2 times frac2theta_z2 pi \
& = R^2 theta_z \
& = R^2 cos^-1(z/R).
endaligned
We can now obtain $A(z)$ by subtracting from $W_z$ the area, $T_z$, of the triangle made out by the lines $OBCO$, which is equal to the area of the rectangle with side lengths equal to those of the line segments $OA$ and $AB$.
Since $A$ has coordinates $(z,0)$ the line segment $OA$ has length $z$; we can calculate the coordinates of $B$ by relying on the formula $x^2 + y^2 = R^2$, which gives coordinates $(z, sqrtR^2 - z^2)$, and so the length of $AB$ is $sqrtR^2 - z^2$. Combining these we get
$$T_z = z sqrtR^2 - z^2$$,
and further more
beginaligned
A(z) &= W_z - T_z \
& = R^2 cos^-1(z/R) - z sqrtR^2 - z^2 \
& = R^2 left( cos^-1(z/R) - (z/R) sqrt1 - (z/R)^2 right),
endaligned
which is as you gave in the question statement.
Calculating $A'(z)$
This now follows from appealing to standard identities for derivatives. If we denote
$$g(x) = cos^-1(x) - x sqrt1 - x^2,$$
then we have $A(z) = R^2 g(z/R)$, and so using the chain rule:
beginaligned
A'(z) & = R^2 fracddz g(z/R) \
& = R^2 g'(z/R) fracddz left(fraczRright) \
& = R g'(z/R).
endaligned
So concentrating on $g(x)$ we differentiate this using the identity for the derivative of $cos^-1$, as well as the product rule to handle the second term to obtain:
$$g'(x) = - frac1sqrt1-x^2 - frac1-2x^2sqrt1-x^2, $$
where the first term above is contributed from $cos^-1$, and the second is from the product term. Rearranging this we have
beginaligned
g'(x) &= -2frac1 - x^2sqrt1-x^2 \
& = -2 sqrt1 - x^2.
endaligned
And so we have
beginaligned
A'(z) &= Rg'(z/R) \
& = -2R sqrt1 - (z/R)^2 \
& = -2 sqrtR^2 - z^2.
endaligned
Substituting this into the original formula provided for $f_Z(z)$ gives the desired answer.
answered Jul 24 at 18:27
owen88
3,2731021
edited Jul 25 at 14:50
answered Jul 24 at 18:27
owen88
3,2731021
answered Jul 24 at 18:27
owen88
3,2731021
answered Jul 24 at 18:27
owen88
3,2731021
3,2731021
Thanks for this brilliant answer. Please could you add some detail on how $A'(z) = R g'(z/R).$ was gotten using the chain rule. I didn't get that part well even with my basic knowledge of the chain rule.
– Abdulhameed
Jul 25 at 0:09
1
I have added a few lines breaking down the chain rule calculation.
– owen88
Jul 25 at 14:50
Those lines are well appreciated and understood. Thank you very much
– Abdulhameed
Jul 26 at 3:49
add a comment |Â
Thanks for this brilliant answer. Please could you add some detail on how $A'(z) = R g'(z/R).$ was gotten using the chain rule. I didn't get that part well even with my basic knowledge of the chain rule.
– Abdulhameed
Jul 25 at 0:09
1
I have added a few lines breaking down the chain rule calculation.
– owen88
Jul 25 at 14:50
Those lines are well appreciated and understood. Thank you very much
– Abdulhameed
Jul 26 at 3:49
Thanks for this brilliant answer. Please could you add some detail on how $A'(z) = R g'(z/R).$ was gotten using the chain rule. I didn't get that part well even with my basic knowledge of the chain rule.
– Abdulhameed
Jul 25 at 0:09
Thanks for this brilliant answer. Please could you add some detail on how $A'(z) = R g'(z/R).$ was gotten using the chain rule. I didn't get that part well even with my basic knowledge of the chain rule.
– Abdulhameed
Jul 25 at 0:09
1
1
I have added a few lines breaking down the chain rule calculation.
– owen88
Jul 25 at 14:50
I have added a few lines breaking down the chain rule calculation.
– owen88
Jul 25 at 14:50
Those lines are well appreciated and understood. Thank you very much
– Abdulhameed
Jul 26 at 3:49
Those lines are well appreciated and understood. Thank you very much
– Abdulhameed
Jul 26 at 3:49
add a comment |Â
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