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Cdf of the area of a stick broken into 2 with UNI(0,1)
Given a stick $L$ broken into 2 with UNI$(0,L)$.
I am trying to find the cdf and pdf of $Y$ which denotes the area of rectangle created from the two lengths.
( one side = $L-p$ other is $p$)
Now, I am confused about how to apply it to the area , as I think the cdf for just the stick length would be close to this:
let $$X: min(p,L-p)implies(X leq y)$$
If $$y geq frac{L}{2} implies P(X leq y)=1$$
$$yleq 0 implies P(X leq y) = 0$$
We conclude:
$$0 leq y leq frac{L}{2} implies P(X leq y)= frac{2y}{L}$$
$$ (X leq y)= (p leq y) cup {p in [L-y, L]}$$
Am I on the right path or should I form a rectangle and try work from scratch from there?
probability probability-theory probability-distributions
asked Nov 20 '18 at 20:13
freddy 19
324
add a comment |
Given a stick $L$ broken into 2 with UNI$(0,L)$.
I am trying to find the cdf and pdf of $Y$ which denotes the area of rectangle created from the two lengths.
( one side = $L-p$ other is $p$)
Now, I am confused about how to apply it to the area , as I think the cdf for just the stick length would be close to this:
let $$X: min(p,L-p)implies(X leq y)$$
If $$y geq frac{L}{2} implies P(X leq y)=1$$
$$yleq 0 implies P(X leq y) = 0$$
We conclude:
$$0 leq y leq frac{L}{2} implies P(X leq y)= frac{2y}{L}$$
$$ (X leq y)= (p leq y) cup {p in [L-y, L]}$$
Am I on the right path or should I form a rectangle and try work from scratch from there?
probability probability-theory probability-distributions
asked Nov 20 '18 at 20:13
freddy 19
324
Use Heron's formula: $A = sqrt{s (s-a)(s-b)(s-c)}$ where $s = {a+b+c over 2}$ is the semiperimeter and $a,b,c$ are the sides (let $a+b+c equiv 1$). Plot admissible regions in an $a$ versus $b$ plot and perform integration.
– David G. Stork
Nov 20 '18 at 20:48
The stick length is $L$, but the break, $p$, is distributed over 0 to 1. (Is that okay or a misprint?)
– Graham Kemp
Nov 20 '18 at 21:07
@Graham Kemp I am sincerely sorry, I corrected it , I meant L .
– freddy 19
Nov 20 '18 at 21:16
add a comment |
Given a stick $L$ broken into 2 with UNI$(0,L)$.
I am trying to find the cdf and pdf of $Y$ which denotes the area of rectangle created from the two lengths.
( one side = $L-p$ other is $p$)
Now, I am confused about how to apply it to the area , as I think the cdf for just the stick length would be close to this:
let $$X: min(p,L-p)implies(X leq y)$$
If $$y geq frac{L}{2} implies P(X leq y)=1$$
$$yleq 0 implies P(X leq y) = 0$$
We conclude:
$$0 leq y leq frac{L}{2} implies P(X leq y)= frac{2y}{L}$$
$$ (X leq y)= (p leq y) cup {p in [L-y, L]}$$
Am I on the right path or should I form a rectangle and try work from scratch from there?
probability probability-theory probability-distributions
asked Nov 20 '18 at 20:13
freddy 19
324
Given a stick $L$ broken into 2 with UNI$(0,L)$.
I am trying to find the cdf and pdf of $Y$ which denotes the area of rectangle created from the two lengths.
( one side = $L-p$ other is $p$)
Now, I am confused about how to apply it to the area , as I think the cdf for just the stick length would be close to this:
let $$X: min(p,L-p)implies(X leq y)$$
If $$y geq frac{L}{2} implies P(X leq y)=1$$
$$yleq 0 implies P(X leq y) = 0$$
We conclude:
$$0 leq y leq frac{L}{2} implies P(X leq y)= frac{2y}{L}$$
$$ (X leq y)= (p leq y) cup {p in [L-y, L]}$$
Am I on the right path or should I form a rectangle and try work from scratch from there?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Nov 20 '18 at 20:13
freddy 19
324
asked Nov 20 '18 at 20:13
freddy 19
324
edited Nov 20 '18 at 21:15
asked Nov 20 '18 at 20:13
freddy 19
324
asked Nov 20 '18 at 20:13
freddy 19
324
asked Nov 20 '18 at 20:13
freddy 19
324
324
Use Heron's formula: $A = sqrt{s (s-a)(s-b)(s-c)}$ where $s = {a+b+c over 2}$ is the semiperimeter and $a,b,c$ are the sides (let $a+b+c equiv 1$). Plot admissible regions in an $a$ versus $b$ plot and perform integration.
– David G. Stork
Nov 20 '18 at 20:48
The stick length is $L$, but the break, $p$, is distributed over 0 to 1. (Is that okay or a misprint?)
– Graham Kemp
Nov 20 '18 at 21:07
@Graham Kemp I am sincerely sorry, I corrected it , I meant L .
– freddy 19
Nov 20 '18 at 21:16
add a comment |
Use Heron's formula: $A = sqrt{s (s-a)(s-b)(s-c)}$ where $s = {a+b+c over 2}$ is the semiperimeter and $a,b,c$ are the sides (let $a+b+c equiv 1$). Plot admissible regions in an $a$ versus $b$ plot and perform integration.
– David G. Stork
Nov 20 '18 at 20:48
The stick length is $L$, but the break, $p$, is distributed over 0 to 1. (Is that okay or a misprint?)
– Graham Kemp
Nov 20 '18 at 21:07
@Graham Kemp I am sincerely sorry, I corrected it , I meant L .
– freddy 19
Nov 20 '18 at 21:16
Use Heron's formula: $A = sqrt{s (s-a)(s-b)(s-c)}$ where $s = {a+b+c over 2}$ is the semiperimeter and $a,b,c$ are the sides (let $a+b+c equiv 1$). Plot admissible regions in an $a$ versus $b$ plot and perform integration.
– David G. Stork
Nov 20 '18 at 20:48
Use Heron's formula: $A = sqrt{s (s-a)(s-b)(s-c)}$ where $s = {a+b+c over 2}$ is the semiperimeter and $a,b,c$ are the sides (let $a+b+c equiv 1$). Plot admissible regions in an $a$ versus $b$ plot and perform integration.
– David G. Stork
Nov 20 '18 at 20:48
The stick length is $L$, but the break, $p$, is distributed over 0 to 1. (Is that okay or a misprint?)
– Graham Kemp
Nov 20 '18 at 21:07
The stick length is $L$, but the break, $p$, is distributed over 0 to 1. (Is that okay or a misprint?)
– Graham Kemp
Nov 20 '18 at 21:07
@Graham Kemp I am sincerely sorry, I corrected it , I meant L .
– freddy 19
Nov 20 '18 at 21:16
@Graham Kemp I am sincerely sorry, I corrected it , I meant L .
– freddy 19
Nov 20 '18 at 21:16
add a comment |
1 Answer
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votes
What you need is the probability distribution of a function of a random variable.
The process is to determine this function and its inverse. E.g. since $f$ is uniform on $[0,1]$, the cdf is $F(X)=X$; $0leq xleq 1$. Then $Y = h(x) = (1-x)(x)$ the area. Then determine the inverse of $h(x)$, call it $v(y)$ (simple algebra). Then
$$F(y) = P(Y < y) = P(x-x^2 < y) = P(x < v(y)) = F(v(y)).$$
See Functions of one random variable
edited Nov 20 '18 at 21:46
rafa11111
1,116417
answered Nov 20 '18 at 21:13
John McGee
1361
Welcome to Math.SE and thank you for your contribution! You might be interested in learning to use Mathjax for typesetting the equations.
– rafa11111
Nov 20 '18 at 21:35
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
What you need is the probability distribution of a function of a random variable.
The process is to determine this function and its inverse. E.g. since $f$ is uniform on $[0,1]$, the cdf is $F(X)=X$; $0leq xleq 1$. Then $Y = h(x) = (1-x)(x)$ the area. Then determine the inverse of $h(x)$, call it $v(y)$ (simple algebra). Then
$$F(y) = P(Y < y) = P(x-x^2 < y) = P(x < v(y)) = F(v(y)).$$
See Functions of one random variable
edited Nov 20 '18 at 21:46
rafa11111
1,116417
answered Nov 20 '18 at 21:13
John McGee
1361
Welcome to Math.SE and thank you for your contribution! You might be interested in learning to use Mathjax for typesetting the equations.
– rafa11111
Nov 20 '18 at 21:35
add a comment |
What you need is the probability distribution of a function of a random variable.
The process is to determine this function and its inverse. E.g. since $f$ is uniform on $[0,1]$, the cdf is $F(X)=X$; $0leq xleq 1$. Then $Y = h(x) = (1-x)(x)$ the area. Then determine the inverse of $h(x)$, call it $v(y)$ (simple algebra). Then
$$F(y) = P(Y < y) = P(x-x^2 < y) = P(x < v(y)) = F(v(y)).$$
See Functions of one random variable
edited Nov 20 '18 at 21:46
rafa11111
1,116417
answered Nov 20 '18 at 21:13
John McGee
1361
Welcome to Math.SE and thank you for your contribution! You might be interested in learning to use Mathjax for typesetting the equations.
– rafa11111
Nov 20 '18 at 21:35
add a comment |
What you need is the probability distribution of a function of a random variable.
The process is to determine this function and its inverse. E.g. since $f$ is uniform on $[0,1]$, the cdf is $F(X)=X$; $0leq xleq 1$. Then $Y = h(x) = (1-x)(x)$ the area. Then determine the inverse of $h(x)$, call it $v(y)$ (simple algebra). Then
$$F(y) = P(Y < y) = P(x-x^2 < y) = P(x < v(y)) = F(v(y)).$$
See Functions of one random variable
edited Nov 20 '18 at 21:46
rafa11111
1,116417
answered Nov 20 '18 at 21:13
John McGee
1361
What you need is the probability distribution of a function of a random variable.
The process is to determine this function and its inverse. E.g. since $f$ is uniform on $[0,1]$, the cdf is $F(X)=X$; $0leq xleq 1$. Then $Y = h(x) = (1-x)(x)$ the area. Then determine the inverse of $h(x)$, call it $v(y)$ (simple algebra). Then
$$F(y) = P(Y < y) = P(x-x^2 < y) = P(x < v(y)) = F(v(y)).$$
See Functions of one random variable
edited Nov 20 '18 at 21:46
rafa11111
1,116417
answered Nov 20 '18 at 21:13
John McGee
1361
edited Nov 20 '18 at 21:46
rafa11111
1,116417
edited Nov 20 '18 at 21:46
rafa11111
1,116417
edited Nov 20 '18 at 21:46
rafa11111
1,116417
1,116417
answered Nov 20 '18 at 21:13
John McGee
1361
answered Nov 20 '18 at 21:13
John McGee
1361
answered Nov 20 '18 at 21:13
John McGee
1361
1361
Welcome to Math.SE and thank you for your contribution! You might be interested in learning to use Mathjax for typesetting the equations.
– rafa11111
Nov 20 '18 at 21:35
add a comment |
Welcome to Math.SE and thank you for your contribution! You might be interested in learning to use Mathjax for typesetting the equations.
– rafa11111
Nov 20 '18 at 21:35
Welcome to Math.SE and thank you for your contribution! You might be interested in learning to use Mathjax for typesetting the equations.
– rafa11111
Nov 20 '18 at 21:35
Welcome to Math.SE and thank you for your contribution! You might be interested in learning to use Mathjax for typesetting the equations.
– rafa11111
Nov 20 '18 at 21:35
add a comment |
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Use Heron's formula: $A = sqrt{s (s-a)(s-b)(s-c)}$ where $s = {a+b+c over 2}$ is the semiperimeter and $a,b,c$ are the sides (let $a+b+c equiv 1$). Plot admissible regions in an $a$ versus $b$ plot and perform integration.
– David G. Stork
Nov 20 '18 at 20:48
The stick length is $L$, but the break, $p$, is distributed over 0 to 1. (Is that okay or a misprint?)
– Graham Kemp
Nov 20 '18 at 21:07
@Graham Kemp I am sincerely sorry, I corrected it , I meant L .
– freddy 19
Nov 20 '18 at 21:16