Question

Find the MLE based on a random sample of size n for the following distribution

N (0,theta)

Answer 1

what have you tried?

Answer 2

I assume you can find the likelihood function.

Answer 3

i really dont even know where to start

Answer 4

do you know the pdf?

Answer 5

\[f(x;\theta)=\frac{1}{\sqrt{2\pi\theta}}e^{-\frac{x^2}{2\theta}}\]

Answer 6

now you need to find the likelihood function

\[L(\theta)=\prod_{i=1}^{n}\frac{1}{\sqrt{2\pi\theta}}\exp\left(-\frac{x_i^2}{2\theta}\right)\]

Answer 7

and how do i go about doing that?

Answer 8

the above is the likelihood function

\[L(\theta):=\prod_{i=1}^{n}f(x_i;\theta)\]

Answer 9

have you done any of these before?

Answer 10

no im soo lost

Answer 11

do you understand my notation?

Answer 12

yea i know what the notations are

Answer 13

the above is the definition of the likelihood function...you then need to take the log of that and then differentiate with respect to \(\theta\)

Answer 14

ayi ayi lol

Answer 15

\[\ell(\theta)=Log(L(\theta))=Log\left(\prod_{i=1}^{n}\frac{1}{\sqrt{2\pi\theta}}\exp\left(-\frac{x_i^2}{2\theta}\right)\right)\]

Answer 16

what does the exp mean?

Answer 17

\[\exp(x)=e^{x}\]

Answer 18

\[=\sum_{i=1}^{n}Log\left(\frac{1}{\sqrt{2\pi\theta}}\exp\left(-\frac{x_i^2}{2\theta}\right)\right)\]

Answer 19

can you simplify this using the properties of logs...then differentiate.

here Log is the natural log

Answer 20

ok yea i was gunna say isnt it the natural log

Answer 21

it doesn't have to be...but it make the simplification easier.

Answer 22

ok.and no im terrible at differentiation

Answer 23

can you simplify it?

Answer 24

\[Log(AB)=Log(A)+Log(B)\]

\[Log(A^r)=r\cdot Log(A)\]

Answer 25

i know log (b)

Answer 26

its just the exponent lol

Answer 27

right?

Answer 28

yes..the Log of e is just the exponent

Answer 29

and log A becomes ?

Answer 30

\[=\sum_{i=1}^{n}\left[Log\left(\frac{1}{\sqrt{2\pi\theta}}\right)+Log\left(\exp\left(-\frac{x_i^2}{2\theta}\right)\right) \right]\]

\[ =\sum_{i=1}^{n}\left[-\frac{1}{2}Log\left(2\pi\theta\right)-\frac{x_i^2}{2\theta} \right]\]

Answer 31

now i have to differentiate that?

Answer 32

\[=-\frac{n}{2}Log\left(2\pi\theta\right)-\frac{1}{2\theta}\sum_{i=1}^{n}x_i^2\]

then differentiate

Answer 33

and of course ihave no idea how to do that..im sorry thank you so much more helping me

Answer 34

treat everything as a constant except \(\theta\)

Answer 35

after taking the derivative you should get

\[-\frac{n}{2\theta}+\frac{1}{2\theta^2}\sum_{i=1}^{n}x_i^2\]

Answer 36

and is that the final answer?

Answer 37

no

Answer 38

what do i have to do next?

Answer 39

set equal to zero and solve for \(\theta\)

Answer 40

ok and that gives you the answer?

Answer 41

it does...though you should really check to see if it is indeed a maximum.

Answer 42

and what if its not?

Answer 43

it will be

Answer 44

i get confused on how to manipulate a function with summations in it

Answer 45

im sorry you prob think im so dumb..i really just dont understand this

Answer 46

so theta equals summation xi2/n?

Answer 47

yes

\[\hat{\theta}=\dfrac{\displaystyle\sum_{i=1}^{n}x_i^2}{n}\]

Answer 48

thank you soo much!!!