calculate variance of unbiased estimator in Rayleigh distribution

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Given : $hattheta = fracsum X_i^22n $ and $ E(X^2) = 2theta $. $hattheta$ is unbiased estimator for $theta$ based on i.i.d sample from $f(x;theta)= fracxthetae^frac-x^22theta$, x < 0, $theta$ < 0. (Rayleigh distribution).



calculate the variance of $hattheta$



$$ Var(hattheta) = Var(fracsum X_i^22n) = frac14n^2nVar(X^2)= frac14nVar(X^2) $$



then we know that the variance can be decomposed:



$$ Var(X) = E [(X - E(X))^2] = E(X^2) - [E(X)]^2 $$



So:



$$ frac14nVar(X^2) = frac 14n [E(X^4)-E(X^2)^2] $$



How can i calculate $E(X^4)$ ?



I am aware that this could be done through integration:



$$E(X^4) = int_0^inf x^4 fracxthetae^frac-x^22theta dx $$



But it seems rather long solution, is there a shortcut?




expectation variance parameter-estimation estimation-theory




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  • Change of variables $u=fracx^22theta$ simplifies the integral.
    – herb steinberg
    Jul 28 at 15:52














up vote
0
down vote

favorite












Given : $hattheta = fracsum X_i^22n $ and $ E(X^2) = 2theta $. $hattheta$ is unbiased estimator for $theta$ based on i.i.d sample from $f(x;theta)= fracxthetae^frac-x^22theta$, x < 0, $theta$ < 0. (Rayleigh distribution).



calculate the variance of $hattheta$



$$ Var(hattheta) = Var(fracsum X_i^22n) = frac14n^2nVar(X^2)= frac14nVar(X^2) $$



then we know that the variance can be decomposed:



$$ Var(X) = E [(X - E(X))^2] = E(X^2) - [E(X)]^2 $$



So:



$$ frac14nVar(X^2) = frac 14n [E(X^4)-E(X^2)^2] $$



How can i calculate $E(X^4)$ ?



I am aware that this could be done through integration:



$$E(X^4) = int_0^inf x^4 fracxthetae^frac-x^22theta dx $$



But it seems rather long solution, is there a shortcut?




expectation variance parameter-estimation estimation-theory




share|cite|improve this question



















  • Change of variables $u=fracx^22theta$ simplifies the integral.
    – herb steinberg
    Jul 28 at 15:52












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given : $hattheta = fracsum X_i^22n $ and $ E(X^2) = 2theta $. $hattheta$ is unbiased estimator for $theta$ based on i.i.d sample from $f(x;theta)= fracxthetae^frac-x^22theta$, x < 0, $theta$ < 0. (Rayleigh distribution).



calculate the variance of $hattheta$



$$ Var(hattheta) = Var(fracsum X_i^22n) = frac14n^2nVar(X^2)= frac14nVar(X^2) $$



then we know that the variance can be decomposed:



$$ Var(X) = E [(X - E(X))^2] = E(X^2) - [E(X)]^2 $$



So:



$$ frac14nVar(X^2) = frac 14n [E(X^4)-E(X^2)^2] $$



How can i calculate $E(X^4)$ ?



I am aware that this could be done through integration:



$$E(X^4) = int_0^inf x^4 fracxthetae^frac-x^22theta dx $$



But it seems rather long solution, is there a shortcut?




expectation variance parameter-estimation estimation-theory




share|cite|improve this question











Given : $hattheta = fracsum X_i^22n $ and $ E(X^2) = 2theta $. $hattheta$ is unbiased estimator for $theta$ based on i.i.d sample from $f(x;theta)= fracxthetae^frac-x^22theta$, x < 0, $theta$ < 0. (Rayleigh distribution).



calculate the variance of $hattheta$



$$ Var(hattheta) = Var(fracsum X_i^22n) = frac14n^2nVar(X^2)= frac14nVar(X^2) $$



then we know that the variance can be decomposed:



$$ Var(X) = E [(X - E(X))^2] = E(X^2) - [E(X)]^2 $$



So:



$$ frac14nVar(X^2) = frac 14n [E(X^4)-E(X^2)^2] $$



How can i calculate $E(X^4)$ ?



I am aware that this could be done through integration:



$$E(X^4) = int_0^inf x^4 fracxthetae^frac-x^22theta dx $$



But it seems rather long solution, is there a shortcut?





expectation variance parameter-estimation estimation-theory





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 28 at 11:06









user1607

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  • Change of variables $u=fracx^22theta$ simplifies the integral.
    – herb steinberg
    Jul 28 at 15:52
















  • Change of variables $u=fracx^22theta$ simplifies the integral.
    – herb steinberg
    Jul 28 at 15:52















Change of variables $u=fracx^22theta$ simplifies the integral.
– herb steinberg
Jul 28 at 15:52




Change of variables $u=fracx^22theta$ simplifies the integral.
– herb steinberg
Jul 28 at 15:52















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