# random sample MLE question

• Sep 9th 2008, 05:20 PM
wik_chick88
random sample MLE question
Given that X1 = 2.3, X2 = 1.9 and X3 = 4.6 is a random sample from
fX(x) = x^3e^(-x/a)
6a^4
for x greater than or equal to 0, calculate the MLE for a.

also, can anyone tell me how to copy and paste formulas such as the one above into the question on this website? i hate how messy it looks like this...thanks
• Sep 9th 2008, 08:54 PM
mr fantastic
Quote:

Originally Posted by wik_chick88
Given that X1 = 2.3, X2 = 1.9 and X3 = 4.6 is a random sample from
fX(x) = x^3e^(-x/a)
6a^4
for x greater than or equal to 0, calculate the MLE for a.

also, can anyone tell me how to copy and paste formulas such as the one above into the question on this website? i hate how messy it looks like this...thanks

The likelihood function is $\displaystyle f(x_1, \, x_2, \, x_3) = \frac{x_1^3 \, x_2^3 \, x_3^3}{6^3 a^{12}} \, e^{-\frac{(x_1 + x_2 + x_3)}{a}}$.

Now solve $\displaystyle \frac{\partial f}{\partial a} = 0$ for $\displaystyle a$ in terms of $\displaystyle x_1, \, x_2$ and $\displaystyle x_3$.

I get $\displaystyle a = \frac{x_1 + x_2 + x_3}{12}$ but I'm known to make careless mistakes.

Now substitute $\displaystyle x_1 = 2.3, \, x_2 = 1.9$ and $\displaystyle x_3 = 4.6$.

If you quote any reply you'll see the latex code. For what it's worth, here's the code I used to generate $\displaystyle f(x_1, \, x_2, \, x_3) = \frac{x_1^3 \, x_2^3 \, x_3^3}{6^3 a^{12}} \, e^{-\frac{(x_1 + x_2 + x_3)}{a}}$:

$$f(x_1, \, x_2, \, x_3) = \frac{x_1^3 \, x_2^3 \, x_3^3}{6^3 a^{12}} \, e^{-\frac{(x_1 + x_2 + x_3)}{a}}$$.

No doubt a latex guru will chip in on how to make it look better.
• Sep 10th 2008, 02:32 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
The likelihood function is $\displaystyle f(x_1, \, x_2, \, x_3) = \frac{x_1^3 \, x_2^3 \, x_3^3}{6^3 a^{12}} \, e^{-\frac{(x_1 + x_2 + x_3)}{a}}$.

Now solve $\displaystyle \frac{\partial f}{\partial a} = 0$ for $\displaystyle a$ in terms of $\displaystyle x_1, \, x_2$ and $\displaystyle x_3$.

I get $\displaystyle a = \frac{x_1 + x_2 + x_3}{12}$ but I'm known to make careless mistakes.

Now substitute $\displaystyle x_1 = 2.3, \, x_2 = 1.9$ and $\displaystyle x_3 = 4.6$.
[snip]

I meant to add that you might find it easier to work with the log-likelihood function $\displaystyle L = \ln f(x_1, \, x_2, \, x_3)$.
• Sep 14th 2008, 03:52 PM
wik_chick88
Quote:

Originally Posted by mr fantastic
I meant to add that you might find it easier to work with the log-likelihood function $\displaystyle L = \ln f(x_1, \, x_2, \, x_3)$.

im a bit confused on how to get the partial derivative of f with respect to a in terms of $\displaystyle x_1, \, x_2, \, x_3$ are you able to show me your steps? and also, if i used $\displaystyle L = \ln f(x_1, \, x_2, \, x_3)$ does that mean i work out partial derivative of L with respect to a instead?
• Sep 14th 2008, 05:23 PM
mr fantastic
Quote:

Originally Posted by wik_chick88
im a bit confused on how to get the partial derivative of f with respect to a in terms of $\displaystyle x_1, \, x_2, \, x_3$ are you able to show me your steps? and also, if i used $\displaystyle L = \ln f(x_1, \, x_2, \, x_3)$ does that mean i work out partial derivative of L with respect to a instead?

Yes.

Use the log-likelihood function rather than the likelihood function to make the calculation simple.

Note that using the usal log rules you have:

$\displaystyle L = \text{constants} - 12 \ln a - \frac{x_1 + x_2 + x_3}{a}$.

Finding $\displaystyle \frac{\partial L}{\partial a}$ and solving $\displaystyle \frac{\partial L}{\partial a} = 0$ should be straightforward.
• Sep 14th 2008, 06:38 PM
wik_chick88
Quote:

Originally Posted by mr fantastic
Yes.

Use the log-likelihood function rather than the likelihood function to make the calculation simple.

Note that using the usal log rules you have:

[tex]L = \text{constants} - 12 \ln a - math].

Finding $\displaystyle \frac{\partial L}{\partial a}$ and solving $\displaystyle \frac{\partial L}{\partial a} = 0$ should be straightforward.

ok. after i derive it and solve for $\displaystyle \frac{\partial L}{\partial a} = 0$ i get a = 1/12. but you originally said that you got a = $\displaystyle \frac{x_1 + x_2 + x_3}{12}$. how do i explain how i substituted $\displaystyle x_1 + x_2 + x_3$ back in?
• Sep 14th 2008, 07:47 PM
mr fantastic
Quote:

Originally Posted by wik_chick88
ok. after i derive it and solve for $\displaystyle \frac{\partial L}{\partial a} = 0$ i get a = 1/12. but you originally said that you got a = $\displaystyle \frac{x_1 + x_2 + x_3}{12}$. how do i explain how i substituted $\displaystyle x_1 + x_2 + x_3$ back in?

$\displaystyle \frac{\partial L}{\partial a} = - \frac{12}{a} + \frac{x_1 + x_2 + x_3}{a^2}$.

The derivative of $\displaystyle -\frac{k}{a}$ is not $\displaystyle \frac{1}{a^2}$, it's $\displaystyle \frac{k}{a^2}$. I'm guessing that's the mistake you've made ....