Particle in Sphere
The position for a particle is random, with uniform distribution on the sphere that has its centre in origin and radius equal to 7. Calculate the expected value of the particle's distance from the origin.
Uniform in this sens should mean
$$
begin{cases}
f(x,y,z) = frac{1}{pi}, sqrt{x^2 + y^2 + z^2} leq 7 \
0, otherwise
end{cases}
$$
But how do I calculate the expected value from here?
uniform-distribution expected-value
asked Nov 26 at 22:13
Kristoffer Jerzy Linder
316
add a comment |
The position for a particle is random, with uniform distribution on the sphere that has its centre in origin and radius equal to 7. Calculate the expected value of the particle's distance from the origin.
Uniform in this sens should mean
$$
begin{cases}
f(x,y,z) = frac{1}{pi}, sqrt{x^2 + y^2 + z^2} leq 7 \
0, otherwise
end{cases}
$$
But how do I calculate the expected value from here?
uniform-distribution expected-value
asked Nov 26 at 22:13
Kristoffer Jerzy Linder
316
add a comment |
The position for a particle is random, with uniform distribution on the sphere that has its centre in origin and radius equal to 7. Calculate the expected value of the particle's distance from the origin.
Uniform in this sens should mean
$$
begin{cases}
f(x,y,z) = frac{1}{pi}, sqrt{x^2 + y^2 + z^2} leq 7 \
0, otherwise
end{cases}
$$
But how do I calculate the expected value from here?
uniform-distribution expected-value
asked Nov 26 at 22:13
Kristoffer Jerzy Linder
316
The position for a particle is random, with uniform distribution on the sphere that has its centre in origin and radius equal to 7. Calculate the expected value of the particle's distance from the origin.
Uniform in this sens should mean
$$
begin{cases}
f(x,y,z) = frac{1}{pi}, sqrt{x^2 + y^2 + z^2} leq 7 \
0, otherwise
end{cases}
$$
But how do I calculate the expected value from here?
uniform-distribution expected-value
uniform-distribution expected-value
asked Nov 26 at 22:13
Kristoffer Jerzy Linder
316
asked Nov 26 at 22:13
Kristoffer Jerzy Linder
316
asked Nov 26 at 22:13
Kristoffer Jerzy Linder
316
asked Nov 26 at 22:13
Kristoffer Jerzy Linder
316
asked Nov 26 at 22:13
Kristoffer Jerzy Linder
316
316
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
By the way, the density should be $f(x,y,z) = frac{1}{frac{4}{3} pi cdot 7^3}$ when $sqrt{x^2+y^2+z^2} le 7$.
If $R$ is the random distance, then the tail sum formula yields
$$E[R] = int_0^infty P(R ge r) , dr.$$
What is $P(R ge r)$?
$$P(R ge r) = begin{cases}0 & r > 7 \ 1 - frac{r^3}{7^3} & 0 le r le 7end{cases}$$
answered Nov 26 at 22:20
angryavian
38.7k23180
But wait, are you sure? Because $$ int_0^1bigg (1-frac{r^3}{7^3}bigg ) dr = 0.99927 $$ and the answer should be around 5
– Kristoffer Jerzy Linder
Nov 26 at 23:24
@KristofferJerzyLinder Fixed a typo
– angryavian
Nov 26 at 23:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
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active
oldest
votes
By the way, the density should be $f(x,y,z) = frac{1}{frac{4}{3} pi cdot 7^3}$ when $sqrt{x^2+y^2+z^2} le 7$.
If $R$ is the random distance, then the tail sum formula yields
$$E[R] = int_0^infty P(R ge r) , dr.$$
What is $P(R ge r)$?
$$P(R ge r) = begin{cases}0 & r > 7 \ 1 - frac{r^3}{7^3} & 0 le r le 7end{cases}$$
answered Nov 26 at 22:20
angryavian
38.7k23180
But wait, are you sure? Because $$ int_0^1bigg (1-frac{r^3}{7^3}bigg ) dr = 0.99927 $$ and the answer should be around 5
– Kristoffer Jerzy Linder
Nov 26 at 23:24
@KristofferJerzyLinder Fixed a typo
– angryavian
Nov 26 at 23:45
add a comment |
By the way, the density should be $f(x,y,z) = frac{1}{frac{4}{3} pi cdot 7^3}$ when $sqrt{x^2+y^2+z^2} le 7$.
If $R$ is the random distance, then the tail sum formula yields
$$E[R] = int_0^infty P(R ge r) , dr.$$
What is $P(R ge r)$?
$$P(R ge r) = begin{cases}0 & r > 7 \ 1 - frac{r^3}{7^3} & 0 le r le 7end{cases}$$
answered Nov 26 at 22:20
angryavian
38.7k23180
But wait, are you sure? Because $$ int_0^1bigg (1-frac{r^3}{7^3}bigg ) dr = 0.99927 $$ and the answer should be around 5
– Kristoffer Jerzy Linder
Nov 26 at 23:24
@KristofferJerzyLinder Fixed a typo
– angryavian
Nov 26 at 23:45
add a comment |
By the way, the density should be $f(x,y,z) = frac{1}{frac{4}{3} pi cdot 7^3}$ when $sqrt{x^2+y^2+z^2} le 7$.
If $R$ is the random distance, then the tail sum formula yields
$$E[R] = int_0^infty P(R ge r) , dr.$$
What is $P(R ge r)$?
$$P(R ge r) = begin{cases}0 & r > 7 \ 1 - frac{r^3}{7^3} & 0 le r le 7end{cases}$$
answered Nov 26 at 22:20
angryavian
38.7k23180
By the way, the density should be $f(x,y,z) = frac{1}{frac{4}{3} pi cdot 7^3}$ when $sqrt{x^2+y^2+z^2} le 7$.
If $R$ is the random distance, then the tail sum formula yields
$$E[R] = int_0^infty P(R ge r) , dr.$$
What is $P(R ge r)$?
$$P(R ge r) = begin{cases}0 & r > 7 \ 1 - frac{r^3}{7^3} & 0 le r le 7end{cases}$$
answered Nov 26 at 22:20
angryavian
38.7k23180
edited Nov 26 at 23:44
answered Nov 26 at 22:20
angryavian
38.7k23180
answered Nov 26 at 22:20
angryavian
38.7k23180
answered Nov 26 at 22:20
angryavian
38.7k23180
38.7k23180
But wait, are you sure? Because $$ int_0^1bigg (1-frac{r^3}{7^3}bigg ) dr = 0.99927 $$ and the answer should be around 5
– Kristoffer Jerzy Linder
Nov 26 at 23:24
@KristofferJerzyLinder Fixed a typo
– angryavian
Nov 26 at 23:45
add a comment |
But wait, are you sure? Because $$ int_0^1bigg (1-frac{r^3}{7^3}bigg ) dr = 0.99927 $$ and the answer should be around 5
– Kristoffer Jerzy Linder
Nov 26 at 23:24
@KristofferJerzyLinder Fixed a typo
– angryavian
Nov 26 at 23:45
But wait, are you sure? Because $$ int_0^1bigg (1-frac{r^3}{7^3}bigg ) dr = 0.99927 $$ and the answer should be around 5
– Kristoffer Jerzy Linder
Nov 26 at 23:24
But wait, are you sure? Because $$ int_0^1bigg (1-frac{r^3}{7^3}bigg ) dr = 0.99927 $$ and the answer should be around 5
– Kristoffer Jerzy Linder
Nov 26 at 23:24
@KristofferJerzyLinder Fixed a typo
– angryavian
Nov 26 at 23:45
@KristofferJerzyLinder Fixed a typo
– angryavian
Nov 26 at 23:45
add a comment |
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