00:01
Hi, i'm david and i'm here to have you answered your question.
00:04
Now let me bring up your question here.
00:12
In the question here we're going to discuss about the leglude function.
00:16
And let me remind you that if we have the x1 up to the xx, f, and then the leglude function ln of some parameter and the x1 up to the xxx.
00:39
Parameter teta, it's equal to the product of the i goes from 1 up to the end and then times with the density of the xi and also we want to revise about the normal distribution.
00:57
Now if we have the x followed by the normal with the mean and standard deviation and then we will have the density of the x equal to one over the x.
01:10
Square root 2 pi time with the sigma and then exponential of the minus x minus the mean divided by the two standard division square and then in this question here this is square here as well in this question we're given the x1 option the x -n will be iid no more with the mu star and the sigma star and we will will start will be in all and similar star square it will be created in zero.
01:49
And the question asked that you construct the leg -loth function of this data set from x1 up to the xn.
01:56
So therefore we have the ln of the x1 up to the xn and then with the mean and the standard divided square, it will equal to by formula will be the product i goes from 1 to n...