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Consider the following model which may be nonlinear:

$Y_{t} = f (X_{t}, \beta_{0}) + \mu_{t}, \hspace{0.2cm} t=1, ..., T$

If we assume that:

  1. $\mu_{t}$ i.i.d with mean = $0$ and variance $\sigma_{0}^{2} < \infty$

  2. The parameter space $B$ is nonempty and compact

  3. $f (X_{t}, \beta)$ is non random for all $\beta \in B$

  4. $\frac{1}{T}\sum_{t=1}^{T}[f (X_{t}, \beta_{0})-f (X_{t}, \beta)]^2$ converges uniformly to a continuous function $A(\beta)$, which attains its minimum only in $\beta = \beta_{0}$

Let $\hat{\beta_{T}}$ be the OLS estimation for $\beta_{0}$.

Does $\hat{\beta_{T}}$ converges in probability to $\beta_{0}$ when $T\rightarrow \infty$?

I'm trying to solve this problem because our professor stated it as a "recommended excercise". I've tried consulting Amemiya's "Advanced Econometrics" book and other references to get an idea but I'm still stuck on the problem. I would appreciate any help.

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  • $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ – gung Jun 8 '15 at 1:34
  • $\begingroup$ Thanks @gung I'm new at this platform so I didn't know about self-study tag. So far, what I don't understand is why does the problem states that $\beta_{T}$ is the OLS estimator for a model which may be nonlinear. Let say the model is, in fact, nonlinear. Can we estimate its parameter vector by the standard OLS formula $\beta = (X'X)^{-1}X'Y$?. Will it be well defined? $\endgroup$ – Marcus Fermat Jun 8 '15 at 1:44
  • $\begingroup$ have you tried finding a counterexample? $\endgroup$ – Christoph Hanck Jun 8 '15 at 3:24
  • $\begingroup$ @ChristophHanck No I haven´t. Do you mean a counterexample for the main problem? Or for the question I asked in my previous comment? $\endgroup$ – Marcus Fermat Jun 8 '15 at 11:27
  • $\begingroup$ For the main problem. For example, some discrete choice model might fit your assumptions, for whose parameters OLS is generally not consistent. $\endgroup$ – Christoph Hanck Jun 8 '15 at 11:29
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OLS is generally consistent for the linear projection coefficients $\beta^\star$, see e.g. any somewhat advanced econometrics textbook like Hayashi's. In the case of a simple regression without constant (for simplicity), $$ \beta^\star=\frac{E(X\cdot Y)}{E(X^2)} $$ Let us consider the case where $P(X_i=-1)=P(X_i=1)=0.5$ and $Y_i=\exp(X_i\beta)+u_i$, $u_i$ independent of $X_i$. Without having checked carefully, this example should fit your set of assumptions. Then, $$ E(X\cdot Y)=E(X\exp(X\beta))=-0.5\exp(-\beta)+0.5\exp(\beta) $$ and $E(X^2)=1$. So for $\beta=1$, OLS is consistent for $0.5[\exp(\beta)-\exp(-\beta)]\approx1.18$, not 1.

Here is an illustration for a sample size of $n=500$ and 10,000 replications to simulate the resulting distribution, which we see centers around 1.18, not 1.

enter image description here

library(MASS)
reps <- 10000
olse <- rep(NA,reps)

n <- 500
beta <- 1
plimols <- (exp(beta) - exp(-beta))/2

for (i in 1:reps) {
  X <- 2*(runif(n)>.5)-1
  u <- rnorm(n,sd=.1)
  y <- exp(beta*X) + u
  olse[i] <- lm(y~X-1)$coefficients  
}
truehist(olse,col="salmon")
abline(v=plimols,col="gold",lwd=3)
abline(v=beta,col="purple",lwd=3)
mean(olse)
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  • $\begingroup$ Indeed, but OP says that "let $\hat{\beta_{T}}$ be the OLS estimation" - my understanding was that he wants to know what happens if you regress, by OLS, $Y$ on $X$ even though the true relationship is nonlinear. I may have misunderstood his question, of course - maybe @MarcusFermat can clarify? $\endgroup$ – Christoph Hanck Jun 10 '15 at 12:47
  • $\begingroup$ Hm, in that case I would have misunderstood - I suggest I wait to withdraw the answer until this gets confirmed, and then the post should be edited replacing OLS with NLS (see his first comment under the original question, though!). $\endgroup$ – Christoph Hanck Jun 10 '15 at 13:29
  • $\begingroup$ Yes, that does clearly support your interpretation - and indicates the source of the ambiguity is not the OP but the author of the question. Perhaps more information about the source (such as a complete reference, if it's from a book) would be useful, and your answer does partly respond to the issue in the comment. I'll delete my earlier comments. $\endgroup$ – Glen_b Jun 10 '15 at 13:46

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