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I am estimating an likelihood function (a structural model). A part of the likelihood function is that $$ p_t=p_{t-1}k_1+x_t(1-k_1) \quad if \ x_t=1 $$ $$ p_t=p_{t-1}k_2+x_t(1-k_2) \quad if \ x_t=0 $$ Here $x_t$ is the independent binary variable. $p_t$ is a intermediate variable for my likelihood function. The parameter I want to estimate is $p_0$, $k_1$ and $k_2$, and they are constrain to be between 0 and 1. So $p_t$ is also between 0 and 1.

$k_1$ and $k_2$ are basically weights on $p_{t-1}$ relative to $x_t$. These can be re-written as: $$ p_t=p_{t-1}x_tk_1+p_{t-1}(1-x_t)k_2+x_t(k_2-k_1) \quad (1) $$

I want to estimate if there is any asymmetric weight of $x_t$, i.e., test if $k_1=k_2$.

Can $k_1$ and $k_2$ be identified in (1)? If they cannot be identified (1), is it possible to transform the equation so that $k_1$ and $k_2$ can be identified?

Edit: Here is my attempt to see if (1) is identifiable or not: $$p_t(k_1,k_2) = p_t(k_1', k_2')$$ $$ x_t(k_2-k_2' - (k_1-k_1')) + p_{t-1}x_t(k_1-k_1') + p_{t-1}(1-x_t)(k_2-k_2')=0$$

When $x_t$=$p_{t-1}=0$, $k_2$ and $k_1$ cannot be identified. Otherwise, they can. So as long as not all $x_t$ and $p_{t-1}$ are zeros, the model can be identified. Is this correct?

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  • $\begingroup$ This looks like an awkward model because it will be hard to guarantee all fitted $\hat p_t$ also are in the interval $[0,1].$ Consider asking a separate question that informs us about your actual statistical problem, because maybe there's a better approach to solving it than this. $\endgroup$
    – whuber
    Commented Jan 21 at 18:45
  • $\begingroup$ Why $p_t$ can be out of this range? Due to error term? $\endgroup$
    – jasmine
    Commented Jan 21 at 19:49
  • $\begingroup$ The chief problem lies in imposing the constraint. I have added an edit to my answer showing an easy way to deal with it by re-expressing your model in terms of differences of the logits of the $p_t.$ $\endgroup$
    – whuber
    Commented Jan 22 at 14:11

1 Answer 1

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Your model is

$$p_t - 1 = k_1(p_{t-1}-1)$$

when $x_t=1$ and otherwise it is

$$p_t = k_2p_{t-1}.$$

Thus, if you have at least one observation with $x_t=1$ and $p_t-1\ne 0,$ you have a nonzero ordered pair $(u,v)$ where $u = p_{t-1} - 1$ and $v=k_1 u.$ That's a standard regression through the origin and clearly $k_1$ is identifiable.

Similarly, when you have at least one observation with $x_t = 0$ and $p_{t-1}\ne 0,$ you have another standard regression through the origin whose coefficient $k_2$ is identifiable.


Supposing $p_t\ne 0$ and $q_t = 1 - p_t \ne 0$ for all $t,$ it is instructive (and potentially helpful) to define

$$z_t = \log(p_t / q_t) = \log(p_t) - \log(1-p_t) = \operatorname{logit}(p_t).$$

Writing $y_t = z_t - z_{t-1},$ we have

$$y_t = \alpha + \beta x_t$$

where $\alpha = \log k_2$ and $\beta = - \log k_1 - \log k_2 .$ This is a standard regression model for the ordered data pairs $(x_t,y_t).$

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