Recent content by tomasrrd

Great. Thanks for the help :)

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...

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...

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...