Maximum likelihood of a statistical model

[the_dane]

Homework Statement

I look at the distribution $(Y_1,Y_2,...,Y_n)$
where
$Y_i=μ+(1+φ x_i)+ε_i$ where $-1<φ<1$ and $-1<x_i<1$ . x's are known numbers. ε's are independent and normally distributed with mean 0 and variance 1.

I need to find the the maximum likelihood estimator for μ and φ

Homework Equations

The Attempt at a Solution

I get to the Log likelihood funcion: $L(Y_1,Y_2,...,Y_n;μ,φ)=∑μ+(1+φ x_i)+ε_i$
When I differentiate it get:
$d L / dμ =∑1/(μ+(1+φ x_i)+ε_i), d L / dφ =∑x/(μ+(1+φ x_i)+ε_i$. Right?
These equations doesn't allow me to set $d L / dμ=0,d L / dφ=0$ because you can't divide by zero. What is going wrong for me?

 

[Stephen Tashi]

I get to the Log likelihood funcion: $L(Y_1,Y_2,...,Y_n;μ,φ)=∑μ+(1+φ x_i)+ε_i$

Before you get to the log liklihood function, what function are you using for the liklihood function?
if the $Y_i$ represent the sample data, the expression for the liklihood function should have the variables $Y_i$ in it. Otherwise, you'd be doing a computation that ignores the data.

 

[Ray Vickson]

Homework Statement

I look at the distribution $(Y_1,Y_2,...,Y_n)$
where
$Y_i=μ+(1+φ x_i)+ε_i$ where $-1<φ<1$ and $-1<x_i<1$ . x's are known numbers. ε's are independent and normally distributed with mean 0 and variance 1.

I need to find the the maximum likelihood estimator for μ and φ

Homework Equations

The Attempt at a Solution

I get to the Log likelihood funcion: $L(Y_1,Y_2,...,Y_n;μ,φ)=∑μ+(1+φ x_i)+ε_i$
When I differentiate it get:
$d L / dμ =∑1/(μ+(1+φ x_i)+ε_i), d L / dφ =∑x/(μ+(1+φ x_i)+ε_i$. Right?
These equations doesn't allow me to set $d L / dμ=0,d L / dφ=0$ because you can't divide by zero. What is going wrong for me?

This does not look like any likelihood function I have ever seen.

Start with the basics: what is the probability density $f_i(y)$ of the random variable $Y_i$, that is, what is the function $f_i(y)$ in the statement $P(y < Y_i < y + dy) = f_i(y)\, dy ?$ In terms of the density functions $f_1(y_1), f_2(y_2), \ldots, f_n(y_n)$, what is the likelihood of an observed event $\{ Y_1 = y_1, Y_2 =y_2, \ldots, Y_n = y_n\}?$ For given $\{y_i\}$ you want to maximize that likelihood function.

 

[the_dane]

Before you get to the log liklihood function, what function are you using for the liklihood function?
if the $Y_i$ represent the sample data, the expression for the liklihood function should have the variables $Y_i$ in it. Otherwise, you'd be doing a computation that ignores the data.

I use $L(Y_1,...Y_n;μ,φ) =Y_1*Y_2*...*Y_n$

 

[Stephen Tashi]

I use $L(Y_1,...Y_n;μ,φ) =Y_1*Y_2*...*Y_n$

That isn't correct. The problem does not say that $Y_i$ is a random variable with an exponential distribution. The problem says that $\epsilon_i$ is a random variable with a normal distribution.

The only random variable with a known distribution in the equation $Y_i = \mu + (1 + \phi x_i) + \epsilon_i$ is the variable $\epsilon_i$.

The liklihood that $\epsilon_i = \alpha_i$ is $\frac{1}{\sqrt{2\pi}} e^{- \frac{\alpha_i^2}{2}}$

Solve the equation $Y_i = \mu + (1 + \phi x_i) + \alpha_i$ for $\alpha_i$ to express that liklihood in terms of $Y_i$.

 

[Ray Vickson]

I use $L(Y_1,...Y_n;μ,φ) =Y_1*Y_2*...*Y_n$

No: you do not multiply the random variables together; if anything, you multiply their probability distributions.

So, to return to my question in #3: what is (a formula for) the probability density function $f_i(y)$ of the random variable $Y_i?$. Until you can answer that question you will get absolutely nowhere with this problem!

 

[the_dane]

No: you do not multiply the random variables together; if anything, you multiply their probability distributions.

So, to return to my question in #3: what is (a formula for) the probability density function $f_i(y)$ of the random variable $Y_i?$. Until you can answer that question you will get absolutely nowhere with this problem!

Can I get the prob. distribution from the formula of $Y_i$.

 

[Ray Vickson]

Can I get the prob. distribution from the formula of $Y_i$.

Yes, that is exactly what you need to do.