Lengths of randomly broken rods
$begingroup$
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
edited Dec 7 '18 at 22:16
JHF
4,7111026
$endgroup$
add a comment |
$begingroup$
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
edited Dec 7 '18 at 22:16
JHF
4,7111026
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 7 '18 at 22:16
JHF
4,7111026
$endgroup$
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
edited Dec 7 '18 at 22:16
JHF
4,7111026
edited Dec 7 '18 at 22:16
JHF
4,7111026
edited Dec 7 '18 at 22:16
JHF
4,7111026
edited Dec 7 '18 at 22:16
JHF
4,7111026
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 7 '18 at 22:55
JHFJHF
4,7111026
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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$begingroup$
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.
answered Dec 7 '18 at 22:55
JHFJHF
4,7111026
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 7 '18 at 22:55
JHFJHF
4,7111026
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 7 '18 at 22:55
JHFJHF
4,7111026
$endgroup$
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.
answered Dec 7 '18 at 22:55
JHFJHF
4,7111026
answered Dec 7 '18 at 22:55
JHFJHF
4,7111026
answered Dec 7 '18 at 22:55
JHFJHF
4,7111026
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
add a comment |
$begingroup$
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
add a comment |
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Question i is the only question I could do. Got really stuck on the other parts. Any hints of what to do?
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– user624037
Dec 7 '18 at 21:43
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Do you know what a uniform distribution is?
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– Trurl
Dec 7 '18 at 22:20
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Yes, and i'm pretty sure the answer to ii) is a uniform distribution. But I don't understand why
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– user624037
Dec 7 '18 at 22:25