Mixing
moments estimation using Rayleigh distribution
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
consider Rayleigh distribution:
$f(x;theta) = frac1theta e^frac-x^2(2theta)$, x > $0$ and $theta$ > $0$
- Show that $E(X^2) = 2theta$
- On the basis of the proceeding item, construct an unbiased estimator $hattheta_1$ for $theta$ based on i.i.d sample $X_1,...,X_n$ for this model
My solutions:
- I have used substitution by parts and u-substitution:
$E(X^2) = int_0^x x^2 frac1theta e^frac-x^2(2theta) dx = [e^frac-x^22theta(x^2-2theta)]_0^x = e^frac-x^22theta(x^2-2theta) + 2theta $
- our professor provided us with an answer that $hattheta_1 = frac12 frac1n sum_i=1^n chi_i^2 $
Questions:
- What am i doing wrong when calculating the $E(X^2)$
- I am not even sure how to approach this problem. I guess i should be using either MSE or the property that $E(theta)= theta$ for unbiased estimators.
My mathematical background is rather weak, stating the obvious is very welcome.
statistics expectation estimation mean-square-error moment-problem
asked Jul 26 at 13:48
user1607
758
add a comment |Â
up vote
1
down vote
favorite
consider Rayleigh distribution:
$f(x;theta) = frac1theta e^frac-x^2(2theta)$, x > $0$ and $theta$ > $0$
- Show that $E(X^2) = 2theta$
- On the basis of the proceeding item, construct an unbiased estimator $hattheta_1$ for $theta$ based on i.i.d sample $X_1,...,X_n$ for this model
My solutions:
- I have used substitution by parts and u-substitution:
$E(X^2) = int_0^x x^2 frac1theta e^frac-x^2(2theta) dx = [e^frac-x^22theta(x^2-2theta)]_0^x = e^frac-x^22theta(x^2-2theta) + 2theta $
- our professor provided us with an answer that $hattheta_1 = frac12 frac1n sum_i=1^n chi_i^2 $
Questions:
- What am i doing wrong when calculating the $E(X^2)$
- I am not even sure how to approach this problem. I guess i should be using either MSE or the property that $E(theta)= theta$ for unbiased estimators.
My mathematical background is rather weak, stating the obvious is very welcome.
statistics expectation estimation mean-square-error moment-problem
asked Jul 26 at 13:48
user1607
758
Your integration doesn't make sense the way it's written -- you're using the same variable name for a bound integration variable and for a free variable in the upper limit.
– joriki
Jul 26 at 13:53
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
consider Rayleigh distribution:
$f(x;theta) = frac1theta e^frac-x^2(2theta)$, x > $0$ and $theta$ > $0$
- Show that $E(X^2) = 2theta$
- On the basis of the proceeding item, construct an unbiased estimator $hattheta_1$ for $theta$ based on i.i.d sample $X_1,...,X_n$ for this model
My solutions:
- I have used substitution by parts and u-substitution:
$E(X^2) = int_0^x x^2 frac1theta e^frac-x^2(2theta) dx = [e^frac-x^22theta(x^2-2theta)]_0^x = e^frac-x^22theta(x^2-2theta) + 2theta $
- our professor provided us with an answer that $hattheta_1 = frac12 frac1n sum_i=1^n chi_i^2 $
Questions:
- What am i doing wrong when calculating the $E(X^2)$
- I am not even sure how to approach this problem. I guess i should be using either MSE or the property that $E(theta)= theta$ for unbiased estimators.
My mathematical background is rather weak, stating the obvious is very welcome.
statistics expectation estimation mean-square-error moment-problem
asked Jul 26 at 13:48
user1607
758
consider Rayleigh distribution:
$f(x;theta) = frac1theta e^frac-x^2(2theta)$, x > $0$ and $theta$ > $0$
- Show that $E(X^2) = 2theta$
- On the basis of the proceeding item, construct an unbiased estimator $hattheta_1$ for $theta$ based on i.i.d sample $X_1,...,X_n$ for this model
My solutions:
- I have used substitution by parts and u-substitution:
$E(X^2) = int_0^x x^2 frac1theta e^frac-x^2(2theta) dx = [e^frac-x^22theta(x^2-2theta)]_0^x = e^frac-x^22theta(x^2-2theta) + 2theta $
- our professor provided us with an answer that $hattheta_1 = frac12 frac1n sum_i=1^n chi_i^2 $
Questions:
- What am i doing wrong when calculating the $E(X^2)$
- I am not even sure how to approach this problem. I guess i should be using either MSE or the property that $E(theta)= theta$ for unbiased estimators.
My mathematical background is rather weak, stating the obvious is very welcome.
statistics expectation estimation mean-square-error moment-problem
asked Jul 26 at 13:48
user1607
758
asked Jul 26 at 13:48
user1607
758
asked Jul 26 at 13:48
user1607
758
asked Jul 26 at 13:48
user1607
758
758
Your integration doesn't make sense the way it's written -- you're using the same variable name for a bound integration variable and for a free variable in the upper limit.
– joriki
Jul 26 at 13:53
add a comment |Â
Your integration doesn't make sense the way it's written -- you're using the same variable name for a bound integration variable and for a free variable in the upper limit.
– joriki
Jul 26 at 13:53
Your integration doesn't make sense the way it's written -- you're using the same variable name for a bound integration variable and for a free variable in the upper limit.
– joriki
Jul 26 at 13:53
Your integration doesn't make sense the way it's written -- you're using the same variable name for a bound integration variable and for a free variable in the upper limit.
– joriki
Jul 26 at 13:53
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
There are a couple of things going on here. First of all, the Rayleigh distribution is given by $$f(x,theta) = fraccolorredxtheta cdot e^-frac-x^22theta,$$ so there is a factor of $x$ missing in your stated definition. Furthermore, the expect expected value of $X^2$ can be calculated by $$E(X^2) = int_0^colorredinfty y^2 cdot f(y,theta) dy = 2theta.$$ Something went wrong with your integration bounds there. I think it is sometimes good practice to give your integration variable a different name than ones you used before in the same problem to avoid confusion (I used $y$ here instead of $x$).
For the second part, notice now that $frac12 X^2$ is an unbiased estimator for the parameter $theta$, since its expected value is exactly $theta$. Using linearity of the expected value we find that $$hattheta := frac1n sum_i=1^n frac12X_i^2$$ also has an expected value of $theta$, which means that it is an unbiased estimator for $theta$.
answered Jul 26 at 14:23
Pjotr5
1,6131615
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There are a couple of things going on here. First of all, the Rayleigh distribution is given by $$f(x,theta) = fraccolorredxtheta cdot e^-frac-x^22theta,$$ so there is a factor of $x$ missing in your stated definition. Furthermore, the expect expected value of $X^2$ can be calculated by $$E(X^2) = int_0^colorredinfty y^2 cdot f(y,theta) dy = 2theta.$$ Something went wrong with your integration bounds there. I think it is sometimes good practice to give your integration variable a different name than ones you used before in the same problem to avoid confusion (I used $y$ here instead of $x$).
For the second part, notice now that $frac12 X^2$ is an unbiased estimator for the parameter $theta$, since its expected value is exactly $theta$. Using linearity of the expected value we find that $$hattheta := frac1n sum_i=1^n frac12X_i^2$$ also has an expected value of $theta$, which means that it is an unbiased estimator for $theta$.
answered Jul 26 at 14:23
Pjotr5
1,6131615
add a comment |Â
up vote
2
down vote
accepted
There are a couple of things going on here. First of all, the Rayleigh distribution is given by $$f(x,theta) = fraccolorredxtheta cdot e^-frac-x^22theta,$$ so there is a factor of $x$ missing in your stated definition. Furthermore, the expect expected value of $X^2$ can be calculated by $$E(X^2) = int_0^colorredinfty y^2 cdot f(y,theta) dy = 2theta.$$ Something went wrong with your integration bounds there. I think it is sometimes good practice to give your integration variable a different name than ones you used before in the same problem to avoid confusion (I used $y$ here instead of $x$).
For the second part, notice now that $frac12 X^2$ is an unbiased estimator for the parameter $theta$, since its expected value is exactly $theta$. Using linearity of the expected value we find that $$hattheta := frac1n sum_i=1^n frac12X_i^2$$ also has an expected value of $theta$, which means that it is an unbiased estimator for $theta$.
answered Jul 26 at 14:23
Pjotr5
1,6131615
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There are a couple of things going on here. First of all, the Rayleigh distribution is given by $$f(x,theta) = fraccolorredxtheta cdot e^-frac-x^22theta,$$ so there is a factor of $x$ missing in your stated definition. Furthermore, the expect expected value of $X^2$ can be calculated by $$E(X^2) = int_0^colorredinfty y^2 cdot f(y,theta) dy = 2theta.$$ Something went wrong with your integration bounds there. I think it is sometimes good practice to give your integration variable a different name than ones you used before in the same problem to avoid confusion (I used $y$ here instead of $x$).
For the second part, notice now that $frac12 X^2$ is an unbiased estimator for the parameter $theta$, since its expected value is exactly $theta$. Using linearity of the expected value we find that $$hattheta := frac1n sum_i=1^n frac12X_i^2$$ also has an expected value of $theta$, which means that it is an unbiased estimator for $theta$.
answered Jul 26 at 14:23
Pjotr5
1,6131615
There are a couple of things going on here. First of all, the Rayleigh distribution is given by $$f(x,theta) = fraccolorredxtheta cdot e^-frac-x^22theta,$$ so there is a factor of $x$ missing in your stated definition. Furthermore, the expect expected value of $X^2$ can be calculated by $$E(X^2) = int_0^colorredinfty y^2 cdot f(y,theta) dy = 2theta.$$ Something went wrong with your integration bounds there. I think it is sometimes good practice to give your integration variable a different name than ones you used before in the same problem to avoid confusion (I used $y$ here instead of $x$).
For the second part, notice now that $frac12 X^2$ is an unbiased estimator for the parameter $theta$, since its expected value is exactly $theta$. Using linearity of the expected value we find that $$hattheta := frac1n sum_i=1^n frac12X_i^2$$ also has an expected value of $theta$, which means that it is an unbiased estimator for $theta$.
answered Jul 26 at 14:23
Pjotr5
1,6131615
answered Jul 26 at 14:23
Pjotr5
1,6131615
answered Jul 26 at 14:23
Pjotr5
1,6131615
answered Jul 26 at 14:23
Pjotr5
1,6131615
1,6131615
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863415%2fmoments-estimation-using-rayleigh-distribution%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Your integration doesn't make sense the way it's written -- you're using the same variable name for a bound integration variable and for a free variable in the upper limit.
– joriki
Jul 26 at 13:53