Method of Moment, MLE and Information matrix We have $$\mathbb{E}[Y_i| X_i] = β_0 + β_1X_i$$
What would be the Method of Moments estimator and MLE for this model?
 A: Now, you have the score function:
$$U =\begin{pmatrix}
\frac{\partial log(L(β_0, β_1, σ^2)}{\partial \beta_0} \\
\frac{\partial log(L(β_0, β_1, σ^2)}{\partial \beta_1}\\
\frac{\partial log(L(β_0, β_1, σ^2)}{\partial \sigma^2}\\
\end{pmatrix}$$
You would need to compute the information matrix $I$ in this way computing the derivative of $U$:
$$I =\begin{pmatrix}
\frac{\partial log(L(β_0, β_1, σ^2)}{\partial^2 \beta_0} &.&. \\
\frac{\partial log(L(β_0, β_1, σ^2)}{\partial \beta_1\beta_0}&\frac{\partial log(L(β_0, β_1, σ^2)}{\partial^2 \beta_1}&.\\
\frac{\partial log(L(β_0, β_1, σ^2)}{\partial \sigma^2\beta_0}&
\frac{\partial log(L(β_0, β_1, σ^2)}{\partial \sigma^2\beta_1}&\frac{\partial log(L(β_0, β_1, σ^2)}{\partial^2 \sigma^2}\\
\end{pmatrix}$$
The dots are related to symmetry as upper and lower diagonal are equal. After some math, you will end up with this:
$$I=\begin{pmatrix}
-\frac{n}{\sigma^2} &.&. \\
-\frac{\sum_{i=1}^{n}x_i}{\sigma^2}&-\frac{\sum_{i=1}^{n}x_i^2}{\sigma^2}&.\\
-\frac{2}{\sigma^3}\sum_{i=1}^{n}(Y_i − β_0 − β_1x_i)&
-\frac{2}{\sigma^3}\sum_{i=1}^{n}x_i(Y_i-\beta_0-\beta_1x_i)&\frac{n}{\sigma^3}-\frac{2}{\sigma^{5}}\sum_{i=1}^{n}(Y_i − β_0 − β_1x_i)^2=0\\
\end{pmatrix}$$
Which is the information matrix.
