# ECN611 PS3
In the solution manual, we have equation(4)

$$\bar{\theta_i} = \frac{1-\frac{\bar{\theta}_{-i} + \epsilon}{2 \epsilon}}{\frac{\bar{\theta}_{-i} + \epsilon}{2 \epsilon}} -1 $$

After some basic algebra, we can get

$$\bar{\theta_i} =\frac{ - 2\bar{\theta}_{-i} }{\bar{\theta}_{-i} + \epsilon}$$

This is what I used to solve the question, in the bottom of page 3 of my solution:

$$k = \frac{-2 \hat{\theta}_i}{ \hat{\theta}_i +\epsilon} = \frac{-2 \frac{-2k}{k+\epsilon}}{\frac{-2k}{k+\epsilon} + \epsilon}$$

Here I used $k=\bar{\theta}_{-i}$ to avoid the confuse from $\bar{\theta}_{i}$.

For the reader's convenience, I wrote down the algebra calculation below.

$$\bar{\theta_i} = \frac{1-\frac{\bar{\theta}_{-i} + \epsilon}{2 \epsilon}}{\frac{\bar{\theta}_{-i} + \epsilon}{2 \epsilon}} -1 \\ = \frac{1-\frac{\bar{\theta}_{-i} + \epsilon}{2 \epsilon} -\frac{\bar{\theta}_{-i} + \epsilon}{2 \epsilon}}{\frac{\bar{\theta}_{-i} + \epsilon}{2 \epsilon}} \\ = \frac{\frac{ - 2\bar{\theta}_{-i} }{2 \epsilon} }{\frac{\bar{\theta}_{-i} + \epsilon}{2 \epsilon}} \\ = \frac{ - 2\bar{\theta}_{-i} }{\bar{\theta}_{-i} + \epsilon}$$