# 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?

• Where's the original post that was put on hold? If you're asking about old exam questions please add the self-study tag and read its tag wiki, modifying your question to follow the guidelines where necessary. If you post the same post a third time without addressing the problems pointed to with the first post, why would it not also be put on hold? If I could find this post I'd point to what it's asking you to change, but perhaps you deleted it, making it rather hard to fix (which is what you really should have done). – Glen_b Mar 24 '15 at 21:18
• When this one gets put on hold as well (as it probably will if you don't make the necessary changes quickly) -- fix the post -- don't delete and repost it. – Glen_b Mar 24 '15 at 21:26
• stats.stackexchange.com/questions/142737/… Here's the other post, I didn't edit it as I thought it might take too long to be answered that way so I just reposted it. I now realise I shouldn't have done that. – roncook Mar 24 '15 at 21:30
• This was the hold reason: "Self-study questions (including textbook exercises, old exam papers, and homework) that seek to understand the concepts are welcome, but those that demand a solution need to indicate clearly at what step help or advice are needed. For help writing a good self-study question, please visit the meta pages." It explains what to do to make your post acceptable. The bottom of the indicated link indicates local policy with regard to particular urgent statements in your post. I suggest you remove them when you edit. – Glen_b Mar 24 '15 at 21:30
• You've added the tag, but you don't seem to have made the kind of changes that are needed. You need to say more about your take on the problem and identify specific issues you need help with. To get to that point you may need to actually open a textbook or read some course notes related to the issues. The link given in the closure reason shows an example of what your question should look like (I realize you have a somewhat different style of question, but as it indicates the present form of question is not acceptable here) – Glen_b Mar 24 '15 at 21:36

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 $C = bY$?

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 $>95$%.