# Thread: random sample MLE question

1. ## 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

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

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

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

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

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

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

No doubt a latex guru will chip in on how to make it look better.

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

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

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

Now substitute $x_1 = 2.3, \, x_2 = 1.9$ and $x_3 = 4.6$.
[snip]
I meant to add that you might find it easier to work with the log-likelihood function $L = \ln f(x_1, \, x_2, \, x_3)$.

4. Originally Posted by mr fantastic
I meant to add that you might find it easier to work with the log-likelihood function $L = \ln f(x_1, \, x_2, \, x_3)$.
im a bit confused on how to get the partial derivative of f with respect to a in terms of $x_1, \, x_2, \, x_3$ are you able to show me your steps? and also, if i used $L = \ln f(x_1, \, x_2, \, x_3)$ does that mean i work out partial derivative of L with respect to a instead?

5. 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 $x_1, \, x_2, \, x_3$ are you able to show me your steps? and also, if i used $L = \ln f(x_1, \, x_2, \, x_3)$ 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:

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

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

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

7. Originally Posted by wik_chick88
ok. after i derive it and solve for $\frac{\partial L}{\partial a} = 0$ i get a = 1/12. but you originally said that you got a = $\frac{x_1 + x_2 + x_3}{12}$. how do i explain how i substituted $x_1 + x_2 + x_3$ back in?
$\frac{\partial L}{\partial a} = - \frac{12}{a} + \frac{x_1 + x_2 + x_3}{a^2}$.

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