truncated model estimation, on an interval of unobserved variable Y* $Pr[L<Y^*<U]=Pr[Y^*<U]-Pr[Y^*<L]$
$=F^*(U)-F^*(L)$
$lnL_n(\theta)=\sum_{i=1}^nd_iln[F^*(U|x_i,\theta)-F^*(L|x_i,\theta)]$
^is the above likelihood function appropriate for this truncated model (Where Y* is the unobserved variable with observable bounds)? how would I go about estimating beta and sigma^2?
 A: It looks like you're asking for a reasonable hint, so let's deal with the truncation, which is the heart of the problem.
First it's desirable to modify the question: if $L$ and $U$ are "observed" bounds as stated, then the true bounds must be introduced as additional parameters in the model. I don't believe that's what the person who set this question intended you to do (but let us know if that's incorrect). Let's assume these bounds are stipulated independently of any of the observations, so that we may treat them as constants.
Let $\Phi$ be the standard Normal cumulative distribution function and let $\phi$ be its density, $\phi(x) = \exp(-x^2/2)/\sqrt{2\pi}.$ Conditional on the regressor $x,$ the model states that the chance $Y^{*}$ lies between $L$ and $U$ is
$$\Pr(L \le Y^{*} \le U\mid x) = \Phi\left(\frac{U - x^\prime \beta}{\sigma}\right) - \Phi\left(\frac{L - x^\prime \beta}{\sigma}\right).$$
Thus, to obtain a proper probability distribution you have to divide the density of $Y$ by this quantity (this is a direct consequence of the formula for conditional probabilities, where we are conditioning $Y$ on the event $L \le Y^{*}\le U$), giving
$$f(y;\beta,\sigma\mid x) = \frac{1}{\sigma}\frac{\phi\left(\frac{y - x^\prime \beta}{\sigma}\right)}{\Phi\left(\frac{U - x^\prime \beta}{\sigma}\right) - \Phi\left(\frac{L - x^\prime \beta}{\sigma}\right)}.$$
These are the terms out of which you need to construct the likelihood.  You cannot ignore the denominators because they depend on the parameters $(\beta,\sigma).$
