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Given $ U(x) = E(X) + \lambda V(x)$

(1) $$E(x) = \frac{1}{2} \gamma X^2 + \epsilon\sum_{k=1}^N|n_k| + \frac{\eta^*}{\tau}\sum_{k=1}^N n_k^2 $$ where $$ \eta^* = \eta - \frac{1}{2}\gamma \tau $$

(2) $$V(x) = \sigma^2 \sum_{k=1}^N \tau x_k^2 $$

Then the partial derivative of U is:

(3) $$ \frac{\partial U}{\partial x_j} = 2\tau(\lambda\sigma^2x_j - \eta^* \frac{x_{j-1}-2x_j+x_{j+1}}{\tau^2}) $$

for j = 1,....,N-1 Then $\frac{\partial U}{\partial x_j} = 0$ is equivalent to

(4) $$ \frac{1}{\tau^2}(x_{j-1}-2x_j+x_{j+1}) = k^2x_j $$

with $$k^2 = \frac{\lambda\sigma^2}{\eta^*} = \frac{\lambda\sigma^2}{\eta(1-\frac{\gamma \tau}{2\eta})} $$

Can you show how (3) is equivalent to (4)?

Thank you.

user3022875
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1 Answers1

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Well, from $\frac{\partial U}{\partial x_j}=0$, divide by $2\tau$ (assuming $\tau\ne0$?), divide by $\eta^*$, replace $\frac{\lambda\sigma^2}{\eta^*}$ by $k^2$, and add the subtrahend on both sides.

joriki
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  • Hi Joriki can you give the first step? – user3022875 Jun 02 '16 at 17:25
  • @user3022875: I did, the first step is to divide by $2\tau$. Or do you mean the step where you write

    $$ 2\tau\left(\lambda\sigma^2x_j - \eta^* \frac{x_{j-1}-2x_j+x_{j+1}}{\tau^2}\right)=0;? $$

     – joriki Jun 02 '16 at 17:48