OLS in terms of means and sample size

Question

Given a model:

$$ y = \beta_0 + \beta_1 \cdot f + u $$

Where $f$ is dummy $=1$ if female and $0$ otherwise, y is height in cm. The sample size is $n_{female}=n_{male}=100 \rightarrow 200$ in total. Further $\bar{y}_{male} = 175$ and $\bar{y}_{female}=165$. Calculate the estimates of parameters.

My attempt:

Using the well know formula:

$$ \boldsymbol{\hat{\beta}} = (\boldsymbol{X}'\boldsymbol{X})^{-1} \boldsymbol{X}'\boldsymbol{y} $$ I get: $$ \begin{bmatrix} 200 & 100 \\ 100 & 100 \\ \end{bmatrix} ^{-1} \begin{bmatrix} 170 \cdot 200 \\ 165 \cdot 200 \end{bmatrix} $$

First the elements in $(\boldsymbol{X}'\boldsymbol{X})^{-1}$, since $X$ is just a bunch of one's, there are 100 female's in the sample and there are 200 males and females in total. For $\boldsymbol{X}'\boldsymbol{y}$, the first element is the "grand mean" of 170, and the second is the just the sample mean of height for females. Both are scaled by 200, since I did not "down-scale" $(\boldsymbol{X}'\boldsymbol{X})^{-1}$.

Is the correct? I ask, because the solution (when multiplying) results in some (very) odd numbers.

Answer 1

The approach is correct, but there's a slight numerical error: there are only $100$ females, not $200$. The mean heights for males and females can be converted to sums via

$$\text{Sum of male heights} = 100 \times 175$$

and

$$\text{Sum of female heights} = 100 \times 165.$$

Therefore the sum of all heights is

$$\text{Sum of all heights} = 100 \times 175 + 100\times 165 = 200 \times 170,$$

as indicated in the question. Consequently the Normal equations are

$$\pmatrix{200 & 100 \\ 100 & 100}\pmatrix{\hat\beta_0 \\ \hat\beta_1} = \pmatrix{200\cdot170 \\ 100 \cdot165}$$

(not $165\cdot 200$ on the right side), with solution

$$(\hat\beta_0, \hat\beta_1) = (175, -10).$$

Answer 2

Im quite confused. What does $u$ mean? Are these residuals? If so, then

$\mathbf{X'X}$ = $\begin{bmatrix} 200 & 100 \\ 100 & 100 \end{bmatrix}$

since

$\mathbf{X} = \frac{\partial{y}}{\partial\beta} = \left[\begin{array}{cccc|cccc} \frac{\partial{y_1}}{\partial\beta_1} & \frac{\partial{y_2}}{\partial\beta_1} & ... & \frac{\partial{y_{n_f}}}{\partial\beta_1} & \frac{\partial{y_{{n_f}+1}}}{\partial\beta_1} & \frac{\partial{y_{{n_f}+2}}}{\partial\beta_1} & ... & \frac{\partial{y_{n_{{n_f}+{n_m}}}}}{\partial\beta_1} \\ \frac{\partial{y_1}}{\partial\beta_2} & \frac{\partial{y_2}}{\partial\beta_2} & ... & \frac{\partial{y_{n_f}}}{\partial\beta_2} & \frac{\partial{y_{{n_f}+1}}}{\partial\beta_2} & \frac{\partial{y_{{n_f}+2}}}{\partial\beta_2} & ... & \frac{\partial{y_{n_{{n_f}+{n_m}}}}}{\partial\beta_2} \\ \end{array}\right]^T$

=

$\left[\begin{array}{cccc|cccc} 1 & 1 & ... & 1 & 1 & 1 & ... & 1 \\ 0 & 0 & ... & 0 & 1 & 1 & ... & 1 \\ \end{array}\right]^T$

Some thoughts:

Given your equation $\beta_1$ IMHO should be 175 and $\beta_2$ = -10. So for the male and female part you get:

$f_m = 175 (+) -10 \times 0 + u = 175 + u$

$f_f = 175 (+) -10 \times 1 + u = 165 + u$

Since you can use

$\mathbf{\beta} = \left(X'X\right)^{-1}X^{T}\mathbf{y}$

to solve for $\beta$ by using the Moore-Penrose Pseudoinverse.

$\left(\left(X'X\right)^{-1}X^{T}\right)^{+}\beta=\left(\left(X'X\right)^{-1}X^{T}\right)^{+}\begin{bmatrix} 175 \\ -10 \end{bmatrix}=\mathbf{y}$

Now $\mathbf{y}$ contains:

$\mathbf{y} \approx \begin{bmatrix} 165_{f_1} & 165_{f_2} & ... 165_{f_{100}} & 175_{m_1} & 175_{m_2} & ... 175_{m_{100}} \end{bmatrix}^T$

Hope it helps!