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POdifference equations in MATLAB - why the need to switch signs?
 

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  1. POdifference equations in MATLAB - why the need to switch signs?
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    <p>Perhaps this is more of a math question than a MATLAB one, not really sure. I'm using MATLAB to compute an economic model - the New Hybrid ISLM model - and there's a confusing step where the author switches the sign of the solution.</p> <p>First, the author declares symbolic variables and sets up a system of difference equations. Note that the suffixes "a" and "2t" both mean "time t+1", "2a" means "time t+2" and "t" means "time t":</p> <pre><code> %% --------------------------[2] MODEL proc-----------------------------%% % Define endogenous vars ('a' denotes t+1 values) syms y2a pi2a ya pia va y2t pi2t yt pit vt ; % Monetary policy rule ia = q1*ya+q2*pia; % ia = q1*(ya-yt)+q2*pia; %%option speed limit policy % Model equations IS = rho*y2a+(1-rho)*yt-sigma*(ia-pi2a)-ya; AS = beta*pi2a+(1-beta)*pit+alpha*ya-pia+va; dum1 = ya-y2t; dum2 = pia-pi2t; MPs = phi*vt-va; optcon = [IS ; AS ; dum1 ; dum2; MPs]; </code></pre> <p>Edit: The equations that are going into the matrix, as they would appear in a textbook are as follows (curly braces indicate time period values, greek letters are parameters):</p> <p>First equation:</p> <pre><code>y{t+1} = rho*y{t+2} + (1-rho)*y{t} - sigma*(i{t+1}-pi{t+2}) </code></pre> <p>Second equation:</p> <pre><code>pi{t+1} = beta*pi{t+2} + (1-beta)*pi{t} + alpha*y{t+1} + v{t+1} </code></pre> <p>Third and fourth are dummies:</p> <pre><code>y{t+1} = y{t+1} pi{t+1} = pi{t+1} </code></pre> <p>Fifth is simple:</p> <pre><code>v{t+1} = phi*v{t} </code></pre> <p>Moving on, the author computes the matrix A: </p> <pre><code> %% ------------------ [3] Linearization proc ------------------------%% % Differentiation xx = [y2a pi2a ya pia va y2t pi2t yt pit vt] ; % define vars jopt = jacobian(optcon,xx); % Define Linear Coefficients coef = eval(jopt); B = [ -coef(:,1:5) ] ; C = [ coef(:,6:10) ] ; % B[c(t+1) l(t+1) k(t+1) z(t+1)] = C[c(t) l(t) k(t) z(t)] A = inv(C)*B ; %(Linearized reduced form ) </code></pre> <p>As far as I understand, this A is the solution to the system. It's the matrix that turns time t+1 and t+2 variables into t and t+1 variables (it's a forward-looking model). My question is essentially why is it necessary to reverse the signs of all the partial derivatives in B in order to get this solution? I'm talking about this step:</p> <pre><code>B = [ -coef(:,1:5) ] ; </code></pre> <p>Reversing the sign here obviously reverses the sign of every component of A, but I don't have a clear understanding of why it's necessary. My apologies if the question is unclear or if this isn't the best place to ask.</p>
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    1. USjefflovejapan
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    1. COCan you supply the equations you're trying to solve in a more readable format? IE, not the matlab code itself but just the model as it would appear in a textbook.
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      1. POdifference equations in MATLAB - why the need to switch signs?
      1. USMarc
    2. COI just updated the post to include the equations, hope that makes it easier to understand. Thanks a lot for taking a look.
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    3. COdo you get different answers if you do not switch sign?
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      1. USAnycorn
 

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