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Consider two non-independent gaussian random variables: $$(t,c)∼\text{BiNormal}[(𝜇𝑡,𝜇𝑐),(𝜎𝑡,𝜎𝑐),𝜌]$$

I'm interested in understanding when (i.e. for which values of the distribution parameters) it's true that $$\frac{\partial E[t |c<f(t)+x]}{\partial x}>0$$ where 

f[t_]:=b*CDF[NormalDistribution[μa, σa], t]

What I've tried:

I've tried generating an analytical expression for the conditional expectation, but this fails (output same as input)

Expectation[t \[Conditioned] c < b*CDF[NormalDistribution[μa, σa], t] + x,
  {t, c} \[Distributed] BinormalDistribution[{μt, μc}, {σt, σc}, ρ]]

I've also tried spelling out the expectation via integrals over the joint distribution, and then taking derivatives. This does produce a valid analytical output, but I cannot reduce the expression (error message "This system cannot be solved with the methods available to Reduce")

D[Integrate[t*PDF[BinormalDistribution[{µt, µc}, {σt, σc}, ρ], {t, c}], 
  {c, -∞, CDF[NormalDistribution[µa, σa], t] + x}, {t, -∞, ∞}]/
  Integrate[PDF[BinormalDistribution[{µt, µc}, {σt, σc}, ρ], {t, c}], 
  {c, -∞, CDF[NormalDistribution[µa, σa], t] + x}, {t, -∞, ∞}], x]

Here is a related question on the analytical treatment of this problem: https://math.stackexchange.com/questions/3692958/parameters-for-which-conditional-expectation-of-non-independent-gaussian-variabl

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    $\begingroup$ Your Expectation code has some errors. There no such BiNormal function. It should be BinormalDistribution. There are parentheses where there should be curly brackets. $\endgroup$
    – JimB
    Commented May 26, 2020 at 22:25
  • $\begingroup$ Thanks, yes. I've fixed the OP with the actual code. What I had written was not an accurate reflection of the code I had tried running (the output doesnt yield error, it simply returns the input) $\endgroup$
    – OO_SE
    Commented May 26, 2020 at 22:42
  • $\begingroup$ You don't want to use the pipe '|' to mean conditioned. Hit escape and type cond then escape, same for 'dist' (distributed). Still, I don't think Mathematica can calculate it for arbitrary parameters with: d = BinormalDistribution[{\[Mu]t, \[Mu]c}, {\[Sigma]t, \[Sigma]c}, \ \[Rho]]; f[t_, b_] := b*CDF[NormalDistribution[\[Mu]a, \[Sigma]a], t]; D[Expectation[ t \[Conditioned] c < f[t, b] + x, {t, c} \[Distributed] d], x] $\endgroup$
    – flinty
    Commented May 26, 2020 at 23:04
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    $\begingroup$ @flinty I think the cut-and-paste issue was that the code was inbetween a numbered list. At least I couldn't get that to show properly. I edited out the numbered list and hopefully didn't mess up the code. And I agree, I don't think there's a nice symbolic solution for the particular $f(t)$ chosen. $\endgroup$
    – JimB
    Commented May 26, 2020 at 23:39
  • $\begingroup$ @JimB ah, perfect. Yes was struggling with the copy paste. Thanks for fixing it $\endgroup$
    – OO_SE
    Commented May 26, 2020 at 23:42

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