Lengths of randomly broken rods












1












$begingroup$


A continuous random variable $X$ has probability density function $f(x)$. The probability that $X leq x$ is given by the function $F(x)$.



i) Explain why $F'(x) = f(x)$.



ii) A rod of length $2a$ is broken into two parts at a point whose position is random. State the form of the probability distribution of the length of the smaller part.



iii) Find the mean value of this length.



iv) Two equal rods, each of length $2a$, are broken into two parts at points whose positions are random. $X$ is the length of the shortest of the four parts thus obtained. Find the probability, $F(x)$, that $X leq x$, where $0 < x leq a$.



v) Hence or otherwise, show that the probability density function of $X$ is given by
$$f(x) = begin{cases} (2(a-x))/a^2 & 0 < x leq a, \ 0 & x leq 0 text{ or } x > a. end{cases}$$



vi) Show that the mean value of $X$ is $a/3$.



vii) Find the mean value of the sum of the two smaller parts and hence show the mean values of the four parts are in the proportions $1:2:4:5$.






probability random-variables







share|cite|improve this question











$endgroup$












  • $begingroup$
    Question i is the only question I could do. Got really stuck on the other parts. Any hints of what to do?
    $endgroup$
    – user624037
    Dec 7 '18 at 21:43










  • $begingroup$
    Do you know what a uniform distribution is?
    $endgroup$
    – Trurl
    Dec 7 '18 at 22:20










  • $begingroup$
    Yes, and i'm pretty sure the answer to ii) is a uniform distribution. But I don't understand why
    $endgroup$
    – user624037
    Dec 7 '18 at 22:25
















1












$begingroup$


A continuous random variable $X$ has probability density function $f(x)$. The probability that $X leq x$ is given by the function $F(x)$.



i) Explain why $F'(x) = f(x)$.



ii) A rod of length $2a$ is broken into two parts at a point whose position is random. State the form of the probability distribution of the length of the smaller part.



iii) Find the mean value of this length.



iv) Two equal rods, each of length $2a$, are broken into two parts at points whose positions are random. $X$ is the length of the shortest of the four parts thus obtained. Find the probability, $F(x)$, that $X leq x$, where $0 < x leq a$.



v) Hence or otherwise, show that the probability density function of $X$ is given by
$$f(x) = begin{cases} (2(a-x))/a^2 & 0 < x leq a, \ 0 & x leq 0 text{ or } x > a. end{cases}$$



vi) Show that the mean value of $X$ is $a/3$.



vii) Find the mean value of the sum of the two smaller parts and hence show the mean values of the four parts are in the proportions $1:2:4:5$.






probability random-variables







share|cite|improve this question











$endgroup$












  • $begingroup$
    Question i is the only question I could do. Got really stuck on the other parts. Any hints of what to do?
    $endgroup$
    – user624037
    Dec 7 '18 at 21:43










  • $begingroup$
    Do you know what a uniform distribution is?
    $endgroup$
    – Trurl
    Dec 7 '18 at 22:20










  • $begingroup$
    Yes, and i'm pretty sure the answer to ii) is a uniform distribution. But I don't understand why
    $endgroup$
    – user624037
    Dec 7 '18 at 22:25














1












1








1


1



$begingroup$


A continuous random variable $X$ has probability density function $f(x)$. The probability that $X leq x$ is given by the function $F(x)$.



i) Explain why $F'(x) = f(x)$.



ii) A rod of length $2a$ is broken into two parts at a point whose position is random. State the form of the probability distribution of the length of the smaller part.



iii) Find the mean value of this length.



iv) Two equal rods, each of length $2a$, are broken into two parts at points whose positions are random. $X$ is the length of the shortest of the four parts thus obtained. Find the probability, $F(x)$, that $X leq x$, where $0 < x leq a$.



v) Hence or otherwise, show that the probability density function of $X$ is given by
$$f(x) = begin{cases} (2(a-x))/a^2 & 0 < x leq a, \ 0 & x leq 0 text{ or } x > a. end{cases}$$



vi) Show that the mean value of $X$ is $a/3$.



vii) Find the mean value of the sum of the two smaller parts and hence show the mean values of the four parts are in the proportions $1:2:4:5$.






probability random-variables







share|cite|improve this question











$endgroup$




A continuous random variable $X$ has probability density function $f(x)$. The probability that $X leq x$ is given by the function $F(x)$.



i) Explain why $F'(x) = f(x)$.



ii) A rod of length $2a$ is broken into two parts at a point whose position is random. State the form of the probability distribution of the length of the smaller part.



iii) Find the mean value of this length.



iv) Two equal rods, each of length $2a$, are broken into two parts at points whose positions are random. $X$ is the length of the shortest of the four parts thus obtained. Find the probability, $F(x)$, that $X leq x$, where $0 < x leq a$.



v) Hence or otherwise, show that the probability density function of $X$ is given by
$$f(x) = begin{cases} (2(a-x))/a^2 & 0 < x leq a, \ 0 & x leq 0 text{ or } x > a. end{cases}$$



vi) Show that the mean value of $X$ is $a/3$.



vii) Find the mean value of the sum of the two smaller parts and hence show the mean values of the four parts are in the proportions $1:2:4:5$.






probability random-variables




probability random-variables






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 22:16









JHF

4,7111026




4,7111026










asked Dec 7 '18 at 21:43







user624037



















  • $begingroup$
    Question i is the only question I could do. Got really stuck on the other parts. Any hints of what to do?
    $endgroup$
    – user624037
    Dec 7 '18 at 21:43










  • $begingroup$
    Do you know what a uniform distribution is?
    $endgroup$
    – Trurl
    Dec 7 '18 at 22:20










  • $begingroup$
    Yes, and i'm pretty sure the answer to ii) is a uniform distribution. But I don't understand why
    $endgroup$
    – user624037
    Dec 7 '18 at 22:25


















  • $begingroup$
    Question i is the only question I could do. Got really stuck on the other parts. Any hints of what to do?
    $endgroup$
    – user624037
    Dec 7 '18 at 21:43










  • $begingroup$
    Do you know what a uniform distribution is?
    $endgroup$
    – Trurl
    Dec 7 '18 at 22:20










  • $begingroup$
    Yes, and i'm pretty sure the answer to ii) is a uniform distribution. But I don't understand why
    $endgroup$
    – user624037
    Dec 7 '18 at 22:25
















$begingroup$
Question i is the only question I could do. Got really stuck on the other parts. Any hints of what to do?
$endgroup$
– user624037
Dec 7 '18 at 21:43




$begingroup$
Question i is the only question I could do. Got really stuck on the other parts. Any hints of what to do?
$endgroup$
– user624037
Dec 7 '18 at 21:43












$begingroup$
Do you know what a uniform distribution is?
$endgroup$
– Trurl
Dec 7 '18 at 22:20




$begingroup$
Do you know what a uniform distribution is?
$endgroup$
– Trurl
Dec 7 '18 at 22:20












$begingroup$
Yes, and i'm pretty sure the answer to ii) is a uniform distribution. But I don't understand why
$endgroup$
– user624037
Dec 7 '18 at 22:25




$begingroup$
Yes, and i'm pretty sure the answer to ii) is a uniform distribution. But I don't understand why
$endgroup$
– user624037
Dec 7 '18 at 22:25










1 Answer
1






active

oldest

votes


















0












$begingroup$

Let $U = U(0,2a)$ be a uniform distribution supported on the interval $(0,2a)$.



For (ii), let $Y$ be the length of the shorter part of the broken rod. Then $Y = min{U, 2a - U}$. So, for $0 < x leq a$,
begin{align}
mathbb{P}(Y leq x) &= mathbb{P}(min{U, 2a - U} leq x) = mathbb{P}(U leq x text{ or } 2a - U leq x)\
&= mathbb{P}(U leq x) + mathbb{P}(2a - U leq x) - mathbb{P}(2a - x leq U leq x) \
&= frac{x}{2a} + frac{x}{2a} - 0 = frac{x}{a}.
end{align}



So $Y sim U(0,a)$ is uniformly distributed on $(0,a)$.



Part (iii) should be standard given part (ii), since if $f_Y(x)$ is the pdf for $Y$, then $$mathbb{E}Y = int_0^a x f_Y(x) , dx,$$ and I'll leave the details to you.



For (iv), let $Y_1$ be the shorter part of the first broken rod and let $Y_2$ be the shorter part of the second broken rod. Then $Y_1$ and $Y_2$ are iid with distribution $Y$, and $X = min{Y_1, Y_2}$. So,
begin{align*}
mathbb{P}(X leq x) &= mathbb{P}(Y_1 leq x text{ or } Y_2 leq x) \
&= mathbb{P}(Y_1 leq x) + mathbb{P}(Y_2 leq x) - mathbb{P}(Y_1 leq x) mathbb{P}(Y_2 leq x) \
&= frac{x}{a} + frac{x}{a} - (frac{x}{a})^2 = frac{2ax-x^2}{a^2}.
end{align*}



Using part (i), you should be able to derive the stated equation. I'll again leave part (v) to you.



Finally, for (vi), use the fact that expectation is additive, so that $mathbb{E}(Y_1 + Y_2) = mathbb{E}Y_1 + mathbb{E}Y_2$, and proceed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay thank you. Am just stuck on the last part. How can I show that the mean values of the four parts are in the proportions 1: 2: 4 : 5?
    $endgroup$
    – user624037
    Dec 7 '18 at 23:46










  • $begingroup$
    Since you know the mean length of the shortest rod out of the four is $frac{a}{3}$, and the mean of the sum of the shortest two from (vii), what is the mean length of the second shortest rod? If you have a broken rod of total length $2a$ and the shorter part has, say, mean length $frac{a}{3}$, what is the mean length of the longer part?
    $endgroup$
    – JHF
    Dec 8 '18 at 23:10











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Let $U = U(0,2a)$ be a uniform distribution supported on the interval $(0,2a)$.



For (ii), let $Y$ be the length of the shorter part of the broken rod. Then $Y = min{U, 2a - U}$. So, for $0 < x leq a$,
begin{align}
mathbb{P}(Y leq x) &= mathbb{P}(min{U, 2a - U} leq x) = mathbb{P}(U leq x text{ or } 2a - U leq x)\
&= mathbb{P}(U leq x) + mathbb{P}(2a - U leq x) - mathbb{P}(2a - x leq U leq x) \
&= frac{x}{2a} + frac{x}{2a} - 0 = frac{x}{a}.
end{align}



So $Y sim U(0,a)$ is uniformly distributed on $(0,a)$.



Part (iii) should be standard given part (ii), since if $f_Y(x)$ is the pdf for $Y$, then $$mathbb{E}Y = int_0^a x f_Y(x) , dx,$$ and I'll leave the details to you.



For (iv), let $Y_1$ be the shorter part of the first broken rod and let $Y_2$ be the shorter part of the second broken rod. Then $Y_1$ and $Y_2$ are iid with distribution $Y$, and $X = min{Y_1, Y_2}$. So,
begin{align*}
mathbb{P}(X leq x) &= mathbb{P}(Y_1 leq x text{ or } Y_2 leq x) \
&= mathbb{P}(Y_1 leq x) + mathbb{P}(Y_2 leq x) - mathbb{P}(Y_1 leq x) mathbb{P}(Y_2 leq x) \
&= frac{x}{a} + frac{x}{a} - (frac{x}{a})^2 = frac{2ax-x^2}{a^2}.
end{align*}



Using part (i), you should be able to derive the stated equation. I'll again leave part (v) to you.



Finally, for (vi), use the fact that expectation is additive, so that $mathbb{E}(Y_1 + Y_2) = mathbb{E}Y_1 + mathbb{E}Y_2$, and proceed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay thank you. Am just stuck on the last part. How can I show that the mean values of the four parts are in the proportions 1: 2: 4 : 5?
    $endgroup$
    – user624037
    Dec 7 '18 at 23:46










  • $begingroup$
    Since you know the mean length of the shortest rod out of the four is $frac{a}{3}$, and the mean of the sum of the shortest two from (vii), what is the mean length of the second shortest rod? If you have a broken rod of total length $2a$ and the shorter part has, say, mean length $frac{a}{3}$, what is the mean length of the longer part?
    $endgroup$
    – JHF
    Dec 8 '18 at 23:10
















0












$begingroup$

Let $U = U(0,2a)$ be a uniform distribution supported on the interval $(0,2a)$.



For (ii), let $Y$ be the length of the shorter part of the broken rod. Then $Y = min{U, 2a - U}$. So, for $0 < x leq a$,
begin{align}
mathbb{P}(Y leq x) &= mathbb{P}(min{U, 2a - U} leq x) = mathbb{P}(U leq x text{ or } 2a - U leq x)\
&= mathbb{P}(U leq x) + mathbb{P}(2a - U leq x) - mathbb{P}(2a - x leq U leq x) \
&= frac{x}{2a} + frac{x}{2a} - 0 = frac{x}{a}.
end{align}



So $Y sim U(0,a)$ is uniformly distributed on $(0,a)$.



Part (iii) should be standard given part (ii), since if $f_Y(x)$ is the pdf for $Y$, then $$mathbb{E}Y = int_0^a x f_Y(x) , dx,$$ and I'll leave the details to you.



For (iv), let $Y_1$ be the shorter part of the first broken rod and let $Y_2$ be the shorter part of the second broken rod. Then $Y_1$ and $Y_2$ are iid with distribution $Y$, and $X = min{Y_1, Y_2}$. So,
begin{align*}
mathbb{P}(X leq x) &= mathbb{P}(Y_1 leq x text{ or } Y_2 leq x) \
&= mathbb{P}(Y_1 leq x) + mathbb{P}(Y_2 leq x) - mathbb{P}(Y_1 leq x) mathbb{P}(Y_2 leq x) \
&= frac{x}{a} + frac{x}{a} - (frac{x}{a})^2 = frac{2ax-x^2}{a^2}.
end{align*}



Using part (i), you should be able to derive the stated equation. I'll again leave part (v) to you.



Finally, for (vi), use the fact that expectation is additive, so that $mathbb{E}(Y_1 + Y_2) = mathbb{E}Y_1 + mathbb{E}Y_2$, and proceed.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Okay thank you. Am just stuck on the last part. How can I show that the mean values of the four parts are in the proportions 1: 2: 4 : 5?
    $endgroup$
    – user624037
    Dec 7 '18 at 23:46










  • $begingroup$
    Since you know the mean length of the shortest rod out of the four is $frac{a}{3}$, and the mean of the sum of the shortest two from (vii), what is the mean length of the second shortest rod? If you have a broken rod of total length $2a$ and the shorter part has, say, mean length $frac{a}{3}$, what is the mean length of the longer part?
    $endgroup$
    – JHF
    Dec 8 '18 at 23:10














0












0








0





$begingroup$

Let $U = U(0,2a)$ be a uniform distribution supported on the interval $(0,2a)$.



For (ii), let $Y$ be the length of the shorter part of the broken rod. Then $Y = min{U, 2a - U}$. So, for $0 < x leq a$,
begin{align}
mathbb{P}(Y leq x) &= mathbb{P}(min{U, 2a - U} leq x) = mathbb{P}(U leq x text{ or } 2a - U leq x)\
&= mathbb{P}(U leq x) + mathbb{P}(2a - U leq x) - mathbb{P}(2a - x leq U leq x) \
&= frac{x}{2a} + frac{x}{2a} - 0 = frac{x}{a}.
end{align}



So $Y sim U(0,a)$ is uniformly distributed on $(0,a)$.



Part (iii) should be standard given part (ii), since if $f_Y(x)$ is the pdf for $Y$, then $$mathbb{E}Y = int_0^a x f_Y(x) , dx,$$ and I'll leave the details to you.



For (iv), let $Y_1$ be the shorter part of the first broken rod and let $Y_2$ be the shorter part of the second broken rod. Then $Y_1$ and $Y_2$ are iid with distribution $Y$, and $X = min{Y_1, Y_2}$. So,
begin{align*}
mathbb{P}(X leq x) &= mathbb{P}(Y_1 leq x text{ or } Y_2 leq x) \
&= mathbb{P}(Y_1 leq x) + mathbb{P}(Y_2 leq x) - mathbb{P}(Y_1 leq x) mathbb{P}(Y_2 leq x) \
&= frac{x}{a} + frac{x}{a} - (frac{x}{a})^2 = frac{2ax-x^2}{a^2}.
end{align*}



Using part (i), you should be able to derive the stated equation. I'll again leave part (v) to you.



Finally, for (vi), use the fact that expectation is additive, so that $mathbb{E}(Y_1 + Y_2) = mathbb{E}Y_1 + mathbb{E}Y_2$, and proceed.






share|cite|improve this answer









$endgroup$



Let $U = U(0,2a)$ be a uniform distribution supported on the interval $(0,2a)$.



For (ii), let $Y$ be the length of the shorter part of the broken rod. Then $Y = min{U, 2a - U}$. So, for $0 < x leq a$,
begin{align}
mathbb{P}(Y leq x) &= mathbb{P}(min{U, 2a - U} leq x) = mathbb{P}(U leq x text{ or } 2a - U leq x)\
&= mathbb{P}(U leq x) + mathbb{P}(2a - U leq x) - mathbb{P}(2a - x leq U leq x) \
&= frac{x}{2a} + frac{x}{2a} - 0 = frac{x}{a}.
end{align}



So $Y sim U(0,a)$ is uniformly distributed on $(0,a)$.



Part (iii) should be standard given part (ii), since if $f_Y(x)$ is the pdf for $Y$, then $$mathbb{E}Y = int_0^a x f_Y(x) , dx,$$ and I'll leave the details to you.



For (iv), let $Y_1$ be the shorter part of the first broken rod and let $Y_2$ be the shorter part of the second broken rod. Then $Y_1$ and $Y_2$ are iid with distribution $Y$, and $X = min{Y_1, Y_2}$. So,
begin{align*}
mathbb{P}(X leq x) &= mathbb{P}(Y_1 leq x text{ or } Y_2 leq x) \
&= mathbb{P}(Y_1 leq x) + mathbb{P}(Y_2 leq x) - mathbb{P}(Y_1 leq x) mathbb{P}(Y_2 leq x) \
&= frac{x}{a} + frac{x}{a} - (frac{x}{a})^2 = frac{2ax-x^2}{a^2}.
end{align*}



Using part (i), you should be able to derive the stated equation. I'll again leave part (v) to you.



Finally, for (vi), use the fact that expectation is additive, so that $mathbb{E}(Y_1 + Y_2) = mathbb{E}Y_1 + mathbb{E}Y_2$, and proceed.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 22:55









JHFJHF

4,7111026




4,7111026












  • $begingroup$
    Okay thank you. Am just stuck on the last part. How can I show that the mean values of the four parts are in the proportions 1: 2: 4 : 5?
    $endgroup$
    – user624037
    Dec 7 '18 at 23:46










  • $begingroup$
    Since you know the mean length of the shortest rod out of the four is $frac{a}{3}$, and the mean of the sum of the shortest two from (vii), what is the mean length of the second shortest rod? If you have a broken rod of total length $2a$ and the shorter part has, say, mean length $frac{a}{3}$, what is the mean length of the longer part?
    $endgroup$
    – JHF
    Dec 8 '18 at 23:10


















  • $begingroup$
    Okay thank you. Am just stuck on the last part. How can I show that the mean values of the four parts are in the proportions 1: 2: 4 : 5?
    $endgroup$
    – user624037
    Dec 7 '18 at 23:46










  • $begingroup$
    Since you know the mean length of the shortest rod out of the four is $frac{a}{3}$, and the mean of the sum of the shortest two from (vii), what is the mean length of the second shortest rod? If you have a broken rod of total length $2a$ and the shorter part has, say, mean length $frac{a}{3}$, what is the mean length of the longer part?
    $endgroup$
    – JHF
    Dec 8 '18 at 23:10
















$begingroup$
Okay thank you. Am just stuck on the last part. How can I show that the mean values of the four parts are in the proportions 1: 2: 4 : 5?
$endgroup$
– user624037
Dec 7 '18 at 23:46




$begingroup$
Okay thank you. Am just stuck on the last part. How can I show that the mean values of the four parts are in the proportions 1: 2: 4 : 5?
$endgroup$
– user624037
Dec 7 '18 at 23:46












$begingroup$
Since you know the mean length of the shortest rod out of the four is $frac{a}{3}$, and the mean of the sum of the shortest two from (vii), what is the mean length of the second shortest rod? If you have a broken rod of total length $2a$ and the shorter part has, say, mean length $frac{a}{3}$, what is the mean length of the longer part?
$endgroup$
– JHF
Dec 8 '18 at 23:10




$begingroup$
Since you know the mean length of the shortest rod out of the four is $frac{a}{3}$, and the mean of the sum of the shortest two from (vii), what is the mean length of the second shortest rod? If you have a broken rod of total length $2a$ and the shorter part has, say, mean length $frac{a}{3}$, what is the mean length of the longer part?
$endgroup$
– JHF
Dec 8 '18 at 23:10


















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mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

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0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
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How to change which sound is reproduced for terminal bell?

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3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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