Method of moments and integration

[semidevil]

i'm learning the method of moments and I'm not understanding on how to do it.

so for example, fy(y, theta) = theta*y^(theta -1). 0<=y<=1.

to find the method of moments estimate for theta.

the book does E(y) = the integeral from 0 to 1 of y * theta*y^(theta-1) dy.

and that becomes theta * y^theta+1/theta + 1

and that becomes theta/ theta + 1...


first question. am I really suppose to integrate y * theta*y^(theta-1) dy?

first of all, the book has never gone to any of the harder integration techniques, and i really have trouble integrating this. how did they do the steps??

 

[juvenal]

i'm learning the method of moments and I'm not understanding on how to do it.

so for example, fy(y, theta) = theta*y^(theta -1). 0<=y<=1.

to find the method of moments estimate for theta.

the book does E(y) = the integeral from 0 to 1 of y * theta*y^(theta-1) dy.

and that becomes theta * y^theta+1/theta + 1

and that becomes theta/ theta + 1...


first question. am I really suppose to integrate y * theta*y^(theta-1) dy?

first of all, the book has never gone to any of the harder integration techniques, and i really have trouble integrating this. how did they do the steps??

Have you learned expected values or means in your class yet? If not - then your class is being badly taught.

The basic formula for calculating the expected value given some pdf is:

Integral[ y f(y) dy]

given a probability distribution function (pdf) f(y). That is equal to the mean, or first moment, and corresponds to the integral you give above.

Second, the integral that you give can be done by someone who knows one month of high-school calculus.

y* theta * y^{theta-1} can be simplified to theta* y^{theta}. Since theta is fixed, the integration is simple. Can you integrate 2*y^2? How about 3*y^3? Same thing.