Loglinearization of balance of payment

General DSGE Modeling
1

Hi everyone,

I’m trying to log-linearize the balance of payment equation in my model. I have read some papers which linearize the BoP equation, but the results of them are different in the coefficient of interest rate. I attach the file that shows my procedure and result, but it is different from others. I don’t know whether it is wrong or not. The confusing point is the deviation form of interest rate. Could you please give me some hints?

Thanks.
Log-linearization .pdf (99.5 KB)

2

In my case I use a Taylor approximation of order 1 around the steady state and the coefficient of \widehat{r}_{t} is r_{ss}, the steps are as follows:

E_{t}B_{t+1}=\left(1+r_{t}\right)B_{t}+\frac{X_{t}}{Q_{t}}-M_{t},

\left(B_{t+1}-B_{ss}\right)=B_{ss}\left(r_{t}-r_{ss}\right)+\left(1+r_{ss}\right)\left(B_{t}-B_{ss}\right)+\frac{1}{Q_{ss}}\left(X_{t}-X_{ss}\right)-\frac{X_{ss}}{Q_{ss}^{2}}\left(Q_{t}-Q_{ss}\right)-\left(M_{t}-M_{ss}\right),

B_{ss}\widehat{B}_{t+1}=B_{ss}r_{ss}\widehat{r}_{t}+\left(1+r_{ss}\right)B_{ss}\widehat{B}_{t}+\frac{X_{ss}}{Q_{ss}}\widehat{X}_{t}-\frac{X_{ss}}{Q_{ss}}\widehat{Q}_{t}-M_{ss}\widehat{M}_{t},

\widehat{B}_{t+1}=r_{ss}\widehat{r}_{t}+\left(1+r_{ss}\right)\widehat{B}_{t}+\frac{1}{B_{ss}}\left[\frac{X_{ss}}{Q_{ss}}\left(\widehat{X}_{t}-Q_{t}\right)-M_{ss}\widehat{M}_{t}\right]

3

Hi Manuel,

Thanks for your reply. The deviation form of net interest rate should be r_hat= r_t-rss instead of r_hat= (r_t-r_ss)/r_ss because net interest rate is a very small value. So the coefficient of r_hat should be 1 in your case. Is that right?

Btw, how do you put maths equations in the text?

Thanks,
Young

4

Yes, you are right, I forget that detail; this will make the coefficient of the net interest rate equal to 1. However, on the other hand, operating in the same way but using the gross interest rate, I get the following:

\widehat{B}_{t+1}=R_{ss}\left(\widehat{R}_{t}+\widehat{B}_{t}\right)+\frac{X_{ss}}{B_{ss}Q_{ss}}\left(\widehat{X}_{t}-\widehat{Q}_{t}\right)-\frac{M_{ss}}{B_{ss}}\widehat{M}_{t}

which is similar to the result you got. In fact,

R_{ss}\widehat{R}_{t} = (1 + r_{ss})(\frac{1 + r_{t} - 1 - r_{ss}}{1 + r_{ss}}) = \widehat{r}_{t}.

Note: To use mathematical expressions, just include the $ sign before and at the end of each expression and use the LaTeX language.

5

Thanks, Manuel. It seems that most of papers I’ve read ignore the detail.