Two-dimensional Maximum likelihood estimates with 2 parameters

699 Views Asked by At

This is the problem in question

So far, I have calculated the logarithmic likelihood function, which I am fairly certain is:

$$L(\lambda_0,\lambda_1) = 4\ln(4)+8\ln(\lambda_1) + \sum_{i=0}^n\left[\ln(x_1^{(i)})+\ln(x_2^{(i)})\right]-\lambda_0\sum_{i=0}^n\left[(x_1^{(i)})^2+(x_2^{(i)})^2\right]$$

However, to maximize $\lambda_1$ and $\lambda_2$, I take the respective partial derivatives and set them to $0$, the results are inconsistent and I do not get a value for either parameters.

Is there something I am doing wrong? Or is it ok to not find solutions to the MLE problem?

1

There are 1 best solutions below

3
On

Guide:

Notice that it is a constrained optimization problem since $\lambda_0$ and $\lambda_1$ are dependent.

Try to find the relationship between $\lambda_1$ and $\lambda_0$,

$$\int_0^\infty \int_0^\infty 4\lambda_1^2x_1x_2 \exp(-\lambda_0 (x_1^2+x_2^2) )\, \, dx_1 \,dx_2=1$$

$$4\lambda_1^2\int_0^\infty x_1 \exp(-\lambda_0x_1^2) \, dx_1 \int_0 ^\infty x_2 \exp(-\lambda_0 x_2^2 )\, dx_2=1$$

$$\left( \lambda_1\int_0^\infty \exp(-\lambda_0x_1^2) (2x_1\, dx_1) \right)\left( \lambda_1\int_0^\infty \exp(-\lambda_0x_2^2) (2x_2\, dx_2) \right)=1$$

$$\left( \lambda_1\int_0^\infty \exp(-\lambda_0x_2^2) (2x_2\, dx_2) \right)^2=1$$