How to find the long-run relationship using this regression (3rd time posted) This is an old exam question I can't figure out.  



I thought to find the relationship you would sum the coefficients on the Ys, so it would be
0.80 + 0.70 - 0.10, giving you C = 1.4Y, but this isn't an option. Could someone tell me where I went wrong? Is it something to do with the fact it's in log form?
INITIAL ANSWER March 24  

Ok. Let's answer this without answering. Your moral obligation to this community, in case it matters to you, is to report back with your work and your answer.

1) In Economics we use the difference of natural logarithms to express (approximately) something specific. It is essentially stated in the body of the exercise.  

2) An estimated relation through regression is essentially an estimation of the expected value, which in turn can be thought of as some long-term average. So don't you think it should validate the assumed long-run values given in the exercise?

3)  Given the assumed long-run values in the exercise what is the relation between $C_{t-1}/Y_{t-1}$ ,  $C_{t}/Y_{t}$, and of $C/Y$, in the long-run? Or between $\Delta \log Y_{t-1}$ and $\Delta \log Y_{t}$ (in light of your answer in 1) above).

4) Given the assumed long-run values in the exercise, do you think that the answer will be of the form $C = a + bY$, or of the form $C = bY$?

The final step will require remembering your answer in 1).

ADDENDUM April 7th
Now that the question came to its proper home, and a two-weeks interval has passed, let's complete this:  

Given the estimated relationship, in order to be consistent with our model, we accept that this relationship will also hold in the long-run. In the long-run some magnitudes are constant. Which ones, in our case? We are told that the long-run growth rate of Consumption and of Income are the same, constant, and equal to $0.25$. This means that in the long-run we will have the equalities

$$C_{t-1}/Y_{t-1}= C_{t}/Y_{t}=C_{LR}/Y_{LR}$$

$$\Delta \log Y_{t-1}=\Delta \log Y_{t} = \Delta \log C_{t} \equiv g =0.25$$

...since the difference of natural logs approximates the growth rate.
Given these, and given the estimated long-run inflation that is given and equal to
$$\Delta \log P_{t} \equiv \pi = 0.1$$
 the long-run relationship is transformed into an equation with a single unknown, $C_t/Y_t$. Specifically, we have, making first symbolic and then numerical substitution:

$$g = 0.8g+0.7g+0.1\ln(C_{LR}/Y_{LR}) -0.15\pi$$

$$\implies g(1-0.8-0.7) + 0.15\pi = 0.1\ln(C_{LR}/Y_{LR})$$

$$\implies  0.25\cdot (-0.5) + 0.15\cdot 0.1 = 0.1\ln(C_{LR}/Y_{LR})$$

$$\implies -0.11 = 0.1\ln(C_{LR}/Y_{LR})$$

At this point, it appears that the "old exam question" made a mistake, because the choices given for answering seem to "ignore" the existence of the coefficient $0.1$ attached to $\ln(C_{LR}/Y_{LR})$. 

If the relation was $-0.11 = \ln(C_{LR}/Y_{LR})$, we would obtain by taking the exponential $e^{-0.11} = C_{LR}/Y_{LR} \implies 0.896 = C_{LR}/Y_{LR} \implies C_{LR} \approx 0.9Y_{LR}$, which is one of the choices given.

But with the coefficient $0.1$ present we obtain

$$... \frac {-0.11}{0.1} = -1.1 =\ln(C_{LR}/Y_{LR}) \implies e^{-1.1} = C_{LR}/Y_{LR} \implies C_{LR}\approx \frac 13Y_{LR}$$

which is not given as an option.

PS: The standard errors of the estimates are not involved in the above calculations. One could comment that their magnitude compared to the magnitude of the estimates indicates that the estimates are "statistically significant" for confidence $&gt;95$%.