I have an equation obtained from inverse logit transformations

$$\left(1+e^{-(\tilde{\alpha}+x_1\tilde{\beta_1})}\right)^{-1} = (1-\phi)\left(1+e^{-(\alpha+x_1\beta_1)}\right)^{-1}+ \phi\left(1+e^{-(\alpha+x_1\beta_1 + \beta_2)}\right)^{-1}$$

I have a numerical value, $\alpha = 0.2, \beta_1 =0.5, \beta_2 =1$ and $\phi = 0.4$

I want to solve for $\tilde\alpha$ and $\tilde\beta_1$

I am new to logistic regression and this is my first time involving this kind of term in my calculation. I am not looking for computed answers, but would appreciate any insight on how to simplify this equation further.

  • $\begingroup$ It is fine to ask self-study questions here, however please note the following: 1) read the tag wiki and always provide some information on what you have done and where you are stuck. 2) take care when entering equations to ensure the preview display is correct (I edited your question to add missing ")" to your exponents). 3) You might consider this, when posting related questions in quick succession. $\endgroup$ – GeoMatt22 Dec 6 '16 at 2:32
  • $\begingroup$ sorry for related questions and I will make sure I dont make mistake next time... I am stuck around this concept for a long time now... could you give me any insight? $\endgroup$ – jayinbluecity Dec 6 '16 at 2:35
  • $\begingroup$ To start, I suggest you update your question with a description of what ideas you have tried. (Append this to the end.) Moreover, your question itself does not appear to fully explain the problem: You appear to have 2 unknowns, but how many equations do you have? From the context of your other question my guess is that $x_1\in\{0,1\}$, but you do not state this. $\endgroup$ – GeoMatt22 Dec 6 '16 at 2:49
  • $\begingroup$ (1) How is this related to logistic regression? (2) Solving for $\tilde\alpha+x_1\tilde\beta_1$ is straightforward: take the reciprocal, subtract $1$, take the log, negate. (3) You cannot uniquely identify either of the unknown variables because you have only one (linear) equation in two unknowns. $\endgroup$ – whuber Dec 6 '16 at 16:28

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