Logistic Regression and utility function

[tomasrrd]

Homework Statement

Hi, I am tryin to solve the following, but I think that something is missing in my understanding.
A company is trying to predict whether a product will be successful. A measurement X (real valued r.v) is made for each product. The company would like to find a threshold $\theta$ so that if $x>=\theta$ the product is supported and otherwise abandoned. Z is Bernoulli variable to model success(1) and failure(0).

The company found that $X\sim Gamma(2.5,0.25)$ and
$$ \Pr(Z=1 | X=x) = \frac {e^{(x-\mu)e^{-\gamma}}}{1+e^{(x-\mu)e^{-\gamma}}}$$

where $\mu, \gamma$ are unknown constants with flat priors.

The utility for the company:
1. product is supported and is success: $b_1=10$
2. the product is supported and if failure: $b_2=-6$
3. the product is abandoned: $b_3=-1$.

One of the questions I am struggling with is:
Find a formula that computes the expected utility for a product given a specific value for $\theta, \gamma, \mu$.

Relevant Equations

NA

I thought that the expected utility is simply the utility of an outcome multiplied by the probability of that outcome. I thought about the following:
set
$$p:=\Pr(Z=1 | X=x) = \frac {e^{(x-\mu)e^{-\gamma}}}{1+e^{(x-\mu)e^{-\gamma}}}$$
and then
$$E(utility)=[b_1⋅p+b_2⋅(1−p)]\Pr(X\leqθ)+b_3⋅\Pr(X>θ)$$
since we have the information thafdt $X∼Gamma(2.5,0.25)$.

The problem is that p itself depends on x, and it doesn't really make sense.

I am not sure what I am missing...

Thanks in advanced.
Tomas.

 

[QuarkyMeson]

It's been a real long time since I've done these, but let me ask you a question. How could utility be the sum of both supporting and abandoning the product? Since you're given the conditions for supporting or abandoning the product, $x \geq 0$ and $x < 0$, shouldn't you have two utility functions depending on the case?

 

[tomasrrd]

Thanks for the answer.

If I understand your remark correctly, I think that what I (kind of) tried to do in the expression $b_1\cdot p + b_2\cdot(1-p)$

But it will still depend of the $x$ value (which depends on the product).

Another idea that I had is that for each $\mu,\gamma$, take a large sample from $Gamma(2.5,0.25)$ and compute an approximation value for that probability. In this case, it will eliminate the dependency on $x$ and we'll get a number that could make sense. But I am not sure that this is the right way.

By the way, there is an error in the equation of the expected utility, and the expressions $\Pr(X\leq\theta)$ and $\Pr(X >\theta)$ should be interchanged (not sure if I can still edit it).

Thank again,
Tomas.

 

[QuarkyMeson]

What I mean is that your utility function U(x) should piecewise, i.e. in the case of abandoning the product the utlitly is just -1, since the probability of you abandoning it is 1. The other case would be based on the probablities that the product is a success or failure.
You could have already done this and I just got thrown by the typo.

You can then write the unconditional expected utility function and take the integral, $E(U(x))=\int_{-\infty}^{\infty}U(x)f(x)dx$, plug everything in and you should get a final function that only depends on $\theta$, $\mu$, $x$, and $\gamma$.

Find a formula that computes the expected utility for a product given a specific value for $\theta, \gamma, \mu$.

Does the problem actually want you to do the numerical integration or does it actually want you to eliminate x? Or just write it in terms of $\theta, \gamma, \mu$ naturally assuming you're going to integrate over x?

 

[tomasrrd]

So, if I understand correctly, it supposed to be as I wrote (just without the typo):

$$ E(U(x))=\int_{-\infty}^{\infty}U(x)f(x)dx = \int_{0}^{\theta}(-1)f_X(x)dx + \int_{\theta}^{\infty}(b_2(1-p) + pb_1)f_X(x)dx$$

Where $f_X$ has the described gamma density?

The next step is to do numerical integration, but I want to understand what I am actually doing.

Thanks again.

 

[QuarkyMeson]

Yeah that looks like right to me.

 

[tomasrrd]

Great. Thanks for the help :)