# OLS estimation for Nonlinear model

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.

• 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. Commented Jun 8, 2015 at 1:34
• 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? Commented Jun 8, 2015 at 1:44
• have you tried finding a counterexample? Commented Jun 8, 2015 at 3:24
• @ChristophHanck No I haven´t. Do you mean a counterexample for the main problem? Or for the question I asked in my previous comment? Commented Jun 8, 2015 at 11:27
• For the main problem. For example, some discrete choice model might fit your assumptions, for whose parameters OLS is generally not consistent. Commented Jun 8, 2015 at 11:29

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.

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)$coefficients } truehist(olse,col="salmon") abline(v=plimols,col="gold",lwd=3) abline(v=beta,col="purple",lwd=3) mean(olse)  • Indeed, but OP says that "let$\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? Commented Jun 10, 2015 at 12:47
• 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!). Commented Jun 10, 2015 at 13:29
• 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. Commented Jun 10, 2015 at 13:46