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Context

Consider the following sequential data generating process for $Y_1$, $Y_2$, $Y_3$. (By sequential I mean that we generate $Y_1$, $Y_2$, $Y_3$ in sequence.):

$Y_1 = X_1^' \beta + \epsilon_1$

$Y_2 = X_2^' \beta + \epsilon_2$

$Y_3 = X_3^' \beta + \epsilon_3$

where,

$X_3 = x_{3,1}$ if $y_1$ $y_2$ > 0,

OR

$X_3 = x_{3,2}$ if $y_1$ $y_2$ < 0,

$\epsilon_i$, ($i$ = 1, 2, 3) are i.i.d $N(0,\sigma^2)$,

$\beta$ is a $p x 1$ vector and

$X_1$, $X_2$, and $X_3$ are vectors of appropriate dimensions.

Question

Suppose we observe the following sequence: {$Y_1$ = $y_1$,$Y_2$ = $y_2$, $X_3$ = $x_{31}$, $Y_3$ = $y_3$} and wish to estimate the parameters $\beta$ and $\sigma$. Is the likelihood function given below the correct function?

L( $\beta$, $\sigma$ | $y_1$, $y_2$, $y_3$, $x_1$, $x_2$, $x_{31}$ ) = ( $f(y_1|x_1,\beta, \sigma)$ $f(y_2|x_2,\beta, \sigma)$ $f(y_3|x_{31},\beta, \sigma)$ ) / Prob( $Y_1 Y_2 >0 $ )

Thanks

EDIT: Fixed some typos and notation in light of comments by Bjørn.

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2 Answers

In order that this be the likelihood function, it is of course necessary that the things that appear to the right of the vertical slash---the $x$s and $y$s---be observed data.

Moreover, it doesn't matter if you multiply by or divide by or entirely omit an expression that doesn't depend on the parameters that are the arguments to the likelihood function, since likelihood functions that differ only by a constant multiple are equivalent ("constant" in this case means not depending on the betas and sigmas that are the arguments to $L$).


Note added a few minutes later: Haste makes waste. I'm going to look at whether that factor does or does not depend on the parameters. If it does, then you will need to be explicit about the way in which it depends on them. But furthermore, if $Y_1$ and $Y_2$ are actually observed, then whether $Y_1 Y_2$ is positive or negative is part of the data, and the likelihood function won't depend on what would have happened but didn't.

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No. The likelihood function is just the probability of observing the $y$'s given the observed $x$'s and the parameters. So you don't need to divide by Prob( $Y_1 Y_2 >0 $ ).

Here is a way to think about it. If the $x$'s were exogenous then you would not divide by the last term. Given the first two observations with $y_1$, $y_2$, $x_1$, and $x_2$, the fact that $y_1 y_2 > 0 $ adds no incremental information.

Finally, suppose you didn't observe the third datum. Then you surely wouldn't divide by the last term.

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