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Bivariate random variables

  1. May 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Suppose that ##(Y_1,Y_2,\ldots,Y_n)## are random variables, where ##Y_i## has an exponential distribution with probability density function ##f_Y(y_i|\theta_i) = \theta_i e^{-\theta_i y_i}##, ##y_i > 0##, ##\theta_i > 0## where ##E(Y_i) = \frac{1}{\theta_i}## and ##\text{Var}(Y_i) = \frac{1}{\theta_i^2}##.
    Suppose that ##\log{(\theta_i)} = \alpha+\beta(z_i-\bar{z})##, where the values ##z_i## are known for ##i=1,2,\ldots,n##.
    We wish to estimate the vector parameter ##\theta = \begin{bmatrix}
    \alpha \\
    \beta
    \end{bmatrix}##.
    1. Show that the log-likelihood function
    $$\ell(\theta) = n\alpha - \sum_{i=1}^{n} y_ie^{\alpha+\beta(z_i-\bar{z})}$$.

    2. Find the two elements of the score statistic ##U(\theta;y)##.
    Write down the two equations which must be solved to find the maximum likelihood estimates ##\hat{\alpha}## and ##\hat{\beta}##.
    Note: Do not attempt to solve these equations.

    3. Show that the information matrix, ##\begin{bmatrix}
    n & 0 \\
    0 & \sum_{i=1}^{n} (z_i-\bar{z})
    \end{bmatrix}##.
    Hint: Recall ##E(Y_i) = \frac{1}{\theta_i}## and ##\theta_i = e^{\alpha+\beta(z_i-\bar{z})}##.

    4. Find the large sample variances of ##\hat{\alpha}## and ##\hat{\beta}## and show that they are asymptotically uncorrelated.

    2. Relevant equations


    3. The attempt at a solution
    1. I think this result we are trying to prove is wrong. Firstly, we have
    $$\begin{split}
    \log{(f_{Y_i}(y_i|\theta_i))} &= \log{\left(\theta_i e^{-\theta_i y_i}\right)} \\
    &= \log{(\theta_i)}-y_i\theta_i \\
    &= \alpha+\beta(z_i-\bar{z})-y_i\theta_i.
    \end{split}$$
    Thus,
    $$\begin{split}
    \ell(\theta) &= \alpha+\beta(z_1-\bar{z})-y_1\theta_1+\alpha+\beta(z_2-\bar{z})-y_2\theta_2+\ldots+\alpha+\beta(z_n-\bar{z})-y_n\theta_n \\
    &= \alpha+\beta(z_1-\bar{z})-y_1e^{\alpha+\beta(z_1-\bar{z})}+\alpha+\beta(z_2-\bar{z})-y_2e^{\alpha+\beta(z_2-\bar{z})}+\ldots+\alpha+\beta(z_n-\bar{z})-y_ne^{\alpha+\beta(z_i-\bar{z})} \\
    &= n\alpha - \sum_{i=1}^{n} y_ie^{\alpha+\beta(z_i-\bar{z})} - \sum_{i=1}^{n} \beta(z_i-\bar{z}).
    \end{split}$$

    2. Firstly,
    $$\begin{split}
    \frac{d\ell}{d\alpha} &= n - \sum_{i=1}^{n} \left(y_ie^{\alpha+\beta(z_i-\bar{z})}\right) \\
    &= n - \sum_{i=1}^{n} y_i\theta_i
    \end{split}$$
    and
    $$\begin{split}
    \frac{d\ell}{d\beta} &= -\sum_{i=1}^{n} (y_i(z_i-\bar{z})e^{\alpha+\beta(z_i-\bar{z})}) - \sum_{i=1}^{n} (z_i-\bar{z}) \\
    &= -\sum_{i=1}^{n} (y_i\theta_i(z_i-\bar{z})) - \sum_{i=1}^{n} (z_i-\bar{z}).
    \end{split}$$
    So the score statistic is
    $$U(\theta;y) = \begin{bmatrix}
    n - \sum_{i=1}^{n} y_i\theta_i \\[1em]
    -\sum_{i=1}^{n} (y_i\theta_i(z_i-\bar{z})) - \sum_{i=1}^{n} (z_i-\bar{z})
    \end{bmatrix}$$
    We find ##\hat{\alpha}## and ##\hat{\beta}## by setting ##U(\theta;y) = 0## and solving the two equations. This means solving ##n - \sum_{i=1}^{n} y_i\theta_i = 0## and ##\sum_{i=1}^{n} (y_i\theta_i(z_i-\bar{z})) + \sum_{i=1}^{n} (z_i-\bar{z}) = 0## for ##\hat{\alpha}## and ##\hat{\beta}## respectively.

    4. I think this result we are trying to prove is wrong. Firstly, we calculate each partial derivative. Doing so gives
    $$\begin{split}
    \frac{\partial^2 \ell}{\partial \alpha^2} &= \frac{\partial}{\partial \alpha}\left(n - \sum_{i=1}^{n} y_i\theta_i\right) \\
    &= -\sum_{i=1}^{n} y_i\theta_i,
    \end{split}$$
    $$\begin{split}
    \frac{\partial^2 \ell}{\partial \beta \partial \alpha} &= \frac{\partial}{\partial \beta} \left(n - \sum_{i=1}^{n} y_i\theta_i\right) \\
    &= -\sum_{i=1}^{n} y_i\theta_i(z_i-\bar{z}) \\
    &= \frac{\partial^2 \ell}{\partial \alpha \partial \beta}
    \end{split}$$
    and
    $$\begin{split}
    \frac{\partial^2 \ell}{\partial \beta^2} &= \frac{\partial}{\partial \beta} \left(-\sum_{i=1}^{n} (y_i\theta_i(z_i-\bar{z})) - \sum_{i=1}^{n} (z_i-\bar{z})\right) \\
    &= -\sum_{i=1}^{n} y_i\theta_i(z_i-\bar{z})^2.
    \end{split}$$
    Thus,
    $$\begin{split}
    I(\theta) &= -E\left(\begin{bmatrix}
    \frac{\partial^2 \ell}{\partial \alpha^2} & \frac{\partial^2 \ell}{\partial \alpha \partial \beta} \\[1em]
    \frac{\partial^2 \ell}{\partial \beta \partial \alpha} & \frac{\partial^2 \ell}{\partial \beta^2}
    \end{bmatrix}\right) \\
    &= E\left(\begin{bmatrix}
    n & \sum_{i=1}^{n} (z_i-\bar{z}) \\[1em]
    \sum_{i=1}^{n} (z_i-\bar{z}) & \sum_{i=1}^{n} (z_i-\bar{z})^2
    \end{bmatrix}\right) \\
    &= \begin{bmatrix}
    n & \sum_{i=1}^{n} (z_i-\bar{z}) \\[1em]
    \sum_{i=1}^{n} (z_i-\bar{z}) & \sum_{i=1}^{n} (z_i-\bar{z})^2
    \end{bmatrix}.
    \end{split}$$
    I'm not sure if these are right. I reckon that the questions are wrong or is it me.
    Please help!!!!
     
  2. jcsd
  3. May 25, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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