I have a model;
$$y_i = \beta_1 + \frac{1}{\beta_2}x_i+\epsilon_i$$
To simplify I use OLS to regress on;
$$y_i = \delta_1 + \delta_2 x_1 + \epsilon_i$$
Thus I obtain the two estimators $\hat{\delta_2}$ and $\hat{\beta_2}=\frac{1}{\hat{\delta_2}}$.
I now want to find whether $\hat{\beta_2}$ is unbiased and/or consistent. And what it's assymptotic distribution is.
NEW ATTEMPT:
We know that:
- For vector-valued continous function $a(.)$ and a vector of random variables $z_n$:
$$plim_{n\rightarrow\infty}a(z_n)= a(plim_{n\rightarrow\infty}z_n)$$
And for our case, with $a(x)=\frac{1}{x}$:
$$plim_{n\rightarrow\infty}a(\hat{\delta_2})= a(plim_{n\rightarrow\infty}\hat{\delta_2}) $$
$$plim_{n\rightarrow\infty}\frac{1}{\hat{\delta_2}}= \frac{1}{plim_{n\rightarrow\infty}\hat{\delta_2}} $$
Because it's OLS we can assume linearity, iid, weak exogeneity ($E[\epsilon_i|x_i]=0$), and the first&second moment is finite and nonsingular [meaning $E[x_i x_i']$ and $E[g_i g_i']$ with $E[x_i *\epsilon_i]=E[g_i]=0$ from weak exogeneity].
And then:
$$plim_{n\rightarrow\infty}\hat{\delta_2}=\delta_2$$
And then, maybe:
$$plim_{n\rightarrow\infty}\hat{\beta_2}=plim_{n\rightarrow\infty}\frac{1}{\hat{\delta_2}}=\frac{1}{plim_{n\rightarrow\infty}\hat{\delta_2}}=\frac{1}{\delta_2}$$
If this was correct, would it show that $\hat{\beta_2}$ is consistent?
And can we say for the asymptotic distribution that:
$$\sqrt(n)(\hat{\delta_2}-\delta_2)\rightarrow_d N(0,E[x_i x_i']^{-1} E[g_i g_i'] E[x_i x_i']^{-1})$$
and then with the delta method:
$$\sqrt(n)(\hat{\beta_2}-\beta_2)\rightarrow_d N(0,E[x_i x_i'] E[g_i g_i'] E[x_i x_i'])$$
But then in the asymptotic distribution of $\beta_2$ there'd still be an $\epsilon_i$ contained, which I don't know how to deal with.
Things I have already thought about:
- [
Since we assume that the model is true. OLS is an unbiased estimator and thus $E[\hat{\theta_2}]=\theta_2$. Can we therefore also say that $E[\hat{\beta_2}]=E[\frac{1}{E[\hat{\theta_2}]}]= E[\frac{1}{\theta_2}] = \frac{1}{\theta_2} =\frac{1}{\frac{1}{\beta_2}}=\beta_2 $ ?]
If not, what am I missing?}
EDIT: I'm thinking this is likely wrong. But I can show, that $\hat{\beta_2}$ is biased using the jensen inequality. (I forgot to add, that we know that $\beta_2>0$.)
If we say that $g(\hat{\theta_2})=\frac{1}{\hat{\theta_2}}$ and g() is convex then:
$$g(E[\hat{\theta_2}])=\frac{1}{E[\hat{\theta_2}]}$$ and $$E[g(\hat{\theta_2})]=E[\frac{1}{\hat{\theta_2}}]=E[\hat{\beta_2}]$$
and thus: $$\frac{1}{E[\hat{\theta_2}]}\leq E[\hat{\beta_2}] \Rightarrow \frac{1}{\theta_2}\leq \beta_2$$.
Might this be right?
- I think, that in order to show, that $\hat{\beta_2}$ is consistent I have to show that $\hat{\beta_2} \rightarrow_p \beta_2$. What steps do I have to pursue to show this?
- If I want to figure out the asymptotic distribution, can I just show $\hat{\beta_2} \rightarrow_p \beta_2$ and thus $\hat{\beta_2} \rightarrow_d \beta_2$?
EDIT: I think that I can only get the asymptotic distribution from the plim, if the plim gives a distribution... but in this case it's a constant.
My alternative attempt thus far:
We know that $$\hat{\theta_2}=(X'X)^{-1}X'y$$ and $$\epsilon|X \sim N(0,\sigma^{2}I_n)$$.
Therefore: $$\hat{\theta_2}|X \sim N(\theta_2,\sigma^{2}(X'X)^{-1})$$
But then I have a $\theta_2$ in the distribution, which I can't relate to $\beta_2$.
Thanks alot for any help, advice, tips or pointers!
Best wishes
