Mixing
Expected value of $sin(x)$
$begingroup$
I have a uniformly distributed random variable $ omega $ in the range $[fracpi2, fracpi{-2}]$. Then I have the function $ s = sin(omega) $ I want to calculate the expected value of this function $ s $.
So far I know that the uniformly distributed random variable can be written as $$ omega = frac1{fracpi2 - - fracpi2} = frac 1pi $$
Then I don't know if the correct way of calculating the expected value is $$ E = int_{-fracpi2}^{fracpi2} frac1pi sin(x) dx $$
or if I'm completely off.
probability statistics expected-value
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 9:22
Michaela HMichaela H
104
$endgroup$
add a comment |
$begingroup$
I have a uniformly distributed random variable $ omega $ in the range $[fracpi2, fracpi{-2}]$. Then I have the function $ s = sin(omega) $ I want to calculate the expected value of this function $ s $.
So far I know that the uniformly distributed random variable can be written as $$ omega = frac1{fracpi2 - - fracpi2} = frac 1pi $$
Then I don't know if the correct way of calculating the expected value is $$ E = int_{-fracpi2}^{fracpi2} frac1pi sin(x) dx $$
or if I'm completely off.
probability statistics expected-value
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 9:22
Michaela HMichaela H
104
$endgroup$
$begingroup$
By antisymmetry, $0$.
$endgroup$
– Yves Daoust
Jan 28 at 9:25
add a comment |
$begingroup$
I have a uniformly distributed random variable $ omega $ in the range $[fracpi2, fracpi{-2}]$. Then I have the function $ s = sin(omega) $ I want to calculate the expected value of this function $ s $.
So far I know that the uniformly distributed random variable can be written as $$ omega = frac1{fracpi2 - - fracpi2} = frac 1pi $$
Then I don't know if the correct way of calculating the expected value is $$ E = int_{-fracpi2}^{fracpi2} frac1pi sin(x) dx $$
or if I'm completely off.
probability statistics expected-value
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 9:22
Michaela HMichaela H
104
$endgroup$
I have a uniformly distributed random variable $ omega $ in the range $[fracpi2, fracpi{-2}]$. Then I have the function $ s = sin(omega) $ I want to calculate the expected value of this function $ s $.
So far I know that the uniformly distributed random variable can be written as $$ omega = frac1{fracpi2 - - fracpi2} = frac 1pi $$
Then I don't know if the correct way of calculating the expected value is $$ E = int_{-fracpi2}^{fracpi2} frac1pi sin(x) dx $$
or if I'm completely off.
probability statistics expected-value
probability statistics expected-value
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 9:22
Michaela HMichaela H
104
edited Jan 28 at 10:16
Jneven
951322
asked Jan 28 at 9:22
Michaela HMichaela H
104
edited Jan 28 at 10:16
Jneven
951322
edited Jan 28 at 10:16
Jneven
951322
edited Jan 28 at 10:16
Jneven
951322
951322
asked Jan 28 at 9:22
Michaela HMichaela H
104
asked Jan 28 at 9:22
Michaela HMichaela H
104
asked Jan 28 at 9:22
Michaela HMichaela H
104
104
$begingroup$
By antisymmetry, $0$.
$endgroup$
– Yves Daoust
Jan 28 at 9:25
add a comment |
$begingroup$
By antisymmetry, $0$.
$endgroup$
– Yves Daoust
Jan 28 at 9:25
$begingroup$
By antisymmetry, $0$.
$endgroup$
– Yves Daoust
Jan 28 at 9:25
$begingroup$
By antisymmetry, $0$.
$endgroup$
– Yves Daoust
Jan 28 at 9:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your working seems fine.
And by the fact that sine is an odd function and $omega$ is uniformly distributed symmetrically about $0$, the integral is evaluated to be $0$.
answered Jan 28 at 9:25
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
The expected value of any random variable $s(omega)$ where $omega$ is having the probability distribution function $f(omega)$ is given by:
$$ E(s(omega)) = int_{-infty}^{infty} s(omega)f(omega)domega$$
since $omega$ is distributed uniformly in the interval $[-pi/2,pi/2]$ we have $$f(omega) = frac{1}{(pi/2-(-pi/2))} = frac{1}{pi}$$
Now, $$E(s(omega))=int_{-infty}^{infty} sin(omega)f(omega)domega$$
or, $$E(s(omega))=int_{-infty}^{infty} sin(omega)frac{1}{pi}domega$$
The limits for $omega$ is from $[-pi/2,pi/2]$ so the integral is
$$E(s(omega))=int_{-pi/2}^{pi/2} sin(omega)frac{1}{pi}domega$$
The answer of this integral is $0$
answered Jan 28 at 10:08
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your working seems fine.
And by the fact that sine is an odd function and $omega$ is uniformly distributed symmetrically about $0$, the integral is evaluated to be $0$.
answered Jan 28 at 9:25
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Your working seems fine.
And by the fact that sine is an odd function and $omega$ is uniformly distributed symmetrically about $0$, the integral is evaluated to be $0$.
answered Jan 28 at 9:25
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Your working seems fine.
And by the fact that sine is an odd function and $omega$ is uniformly distributed symmetrically about $0$, the integral is evaluated to be $0$.
answered Jan 28 at 9:25
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
Your working seems fine.
And by the fact that sine is an odd function and $omega$ is uniformly distributed symmetrically about $0$, the integral is evaluated to be $0$.
answered Jan 28 at 9:25
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 28 at 9:25
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 28 at 9:25
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 28 at 9:25
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
add a comment |
add a comment |
$begingroup$
The expected value of any random variable $s(omega)$ where $omega$ is having the probability distribution function $f(omega)$ is given by:
$$ E(s(omega)) = int_{-infty}^{infty} s(omega)f(omega)domega$$
since $omega$ is distributed uniformly in the interval $[-pi/2,pi/2]$ we have $$f(omega) = frac{1}{(pi/2-(-pi/2))} = frac{1}{pi}$$
Now, $$E(s(omega))=int_{-infty}^{infty} sin(omega)f(omega)domega$$
or, $$E(s(omega))=int_{-infty}^{infty} sin(omega)frac{1}{pi}domega$$
The limits for $omega$ is from $[-pi/2,pi/2]$ so the integral is
$$E(s(omega))=int_{-pi/2}^{pi/2} sin(omega)frac{1}{pi}domega$$
The answer of this integral is $0$
answered Jan 28 at 10:08
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
add a comment |
$begingroup$
The expected value of any random variable $s(omega)$ where $omega$ is having the probability distribution function $f(omega)$ is given by:
$$ E(s(omega)) = int_{-infty}^{infty} s(omega)f(omega)domega$$
since $omega$ is distributed uniformly in the interval $[-pi/2,pi/2]$ we have $$f(omega) = frac{1}{(pi/2-(-pi/2))} = frac{1}{pi}$$
Now, $$E(s(omega))=int_{-infty}^{infty} sin(omega)f(omega)domega$$
or, $$E(s(omega))=int_{-infty}^{infty} sin(omega)frac{1}{pi}domega$$
The limits for $omega$ is from $[-pi/2,pi/2]$ so the integral is
$$E(s(omega))=int_{-pi/2}^{pi/2} sin(omega)frac{1}{pi}domega$$
The answer of this integral is $0$
answered Jan 28 at 10:08
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
add a comment |
$begingroup$
The expected value of any random variable $s(omega)$ where $omega$ is having the probability distribution function $f(omega)$ is given by:
$$ E(s(omega)) = int_{-infty}^{infty} s(omega)f(omega)domega$$
since $omega$ is distributed uniformly in the interval $[-pi/2,pi/2]$ we have $$f(omega) = frac{1}{(pi/2-(-pi/2))} = frac{1}{pi}$$
Now, $$E(s(omega))=int_{-infty}^{infty} sin(omega)f(omega)domega$$
or, $$E(s(omega))=int_{-infty}^{infty} sin(omega)frac{1}{pi}domega$$
The limits for $omega$ is from $[-pi/2,pi/2]$ so the integral is
$$E(s(omega))=int_{-pi/2}^{pi/2} sin(omega)frac{1}{pi}domega$$
The answer of this integral is $0$
answered Jan 28 at 10:08
SNEHIL SANYALSNEHIL SANYAL
654110
$endgroup$
The expected value of any random variable $s(omega)$ where $omega$ is having the probability distribution function $f(omega)$ is given by:
$$ E(s(omega)) = int_{-infty}^{infty} s(omega)f(omega)domega$$
since $omega$ is distributed uniformly in the interval $[-pi/2,pi/2]$ we have $$f(omega) = frac{1}{(pi/2-(-pi/2))} = frac{1}{pi}$$
Now, $$E(s(omega))=int_{-infty}^{infty} sin(omega)f(omega)domega$$
or, $$E(s(omega))=int_{-infty}^{infty} sin(omega)frac{1}{pi}domega$$
The limits for $omega$ is from $[-pi/2,pi/2]$ so the integral is
$$E(s(omega))=int_{-pi/2}^{pi/2} sin(omega)frac{1}{pi}domega$$
The answer of this integral is $0$
answered Jan 28 at 10:08
SNEHIL SANYALSNEHIL SANYAL
654110
answered Jan 28 at 10:08
SNEHIL SANYALSNEHIL SANYAL
654110
answered Jan 28 at 10:08
SNEHIL SANYALSNEHIL SANYAL
654110
answered Jan 28 at 10:08
SNEHIL SANYALSNEHIL SANYAL
654110
654110
add a comment |
add a comment |
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$begingroup$
By antisymmetry, $0$.
$endgroup$
– Yves Daoust
Jan 28 at 9:25