Covariance between f(x+y) and f(x) Is there some theorem that allows us to work out:
$Cov\big( f(x+y), f(x)\big)$
We now the $Var\big(f(x)\big)$ and $Var\big(f(x+y)\big)$ and also the $Var(x)$, $Var(y)$ and the $Covar(x,y)$
Our specific case is the following:
What is the $Covar\big(\Phi(\beta_0+\beta_1),\Phi(\beta_0 )\big)$, with $\beta_0$ and $\beta_1$ probit model coefficients with variance and covariance known.
 A: The simple probit model, with one binary regressor, $X=\{0,1\}$, for an i.i.d. sample of size $n$, and where $n_1$ is the number of observations where $X=1$ and $n_0$ is the number of observations where $X=0$, $n_1+n_0 =n$, is
$$\begin{align} &p_{1|1}=P(Y=1|X=1)=\Phi(\beta_0+\beta_1X)\\
&p_{1|0}=P(Y=1|X=0)=\Phi(\beta_0)\end{align}$$
The first-order conditions that the maximum likelihood estimator must satisfy are
$$\hat \beta_0: \sum_{i=1}^n\left[\frac {\phi(\hat \beta_0+\hat \beta_1x_i)}{\Phi(\hat \beta_0+\hat \beta_1x_i)}y_i -(1-y_i)\frac {\phi(\hat \beta_0+\hat \beta_1x_i)}{1-\Phi(\hat \beta_0+\hat \beta_1x_i)}\right]=0 \qquad [1]$$
$$\hat \beta_1: \sum_{x_i=1}\left[\frac {\phi(\hat \beta_0+\hat \beta_1)}{\Phi(\hat \beta_0+\hat \beta_1)}y_i -(1-y_i)\frac {\phi(\hat \beta_0+\hat \beta_1)}{1-\Phi(\hat \beta_0+\hat \beta_1)}\right]=0 \qquad [2]$$
Subtracting eq. $[2]$ from eq. $[1]$ we also obtain
$$\sum_{x_i=0}\left[\frac {\phi(\hat \beta_0)}{\Phi(\hat \beta_0)}y_i -(1-y_i)\frac {\phi(\hat \beta_0)}{1-\Phi(\hat \beta_0)}\right]=0 \qquad [3]$$
Working eq.$[2]$, we have
$$ \sum_{x_i=1}\left[\frac {\phi(\hat \beta_0+\hat \beta_1)\left(1-\Phi(\hat \beta_0+\hat \beta_1)\right)y_i-(1-y_i)\phi(\hat \beta_0+\hat \beta_1)\Phi(\hat \beta_0+\hat \beta_1)}{\Phi(\hat \beta_0+\hat \beta_1)\left(1-\Phi(\hat \beta_0+\hat \beta_1)\right)} \right]=0$$
$$\Rightarrow \sum_{x_i=1}\left[\frac {\phi(\hat \beta_0+\hat \beta_1)y_i-\phi(\hat \beta_0+\hat \beta_1)\Phi(\hat \beta_0+\hat \beta_1)}{\Phi(\hat \beta_0+\hat \beta_1)\left(1-\Phi(\hat \beta_0+\hat \beta_1)\right)} \right]=0$$
$$\Rightarrow \left[\frac {\phi(\hat \beta_0+\hat \beta_1)}{\Phi(\hat \beta_0+\hat \beta_1)\left(1-\Phi(\hat \beta_0+\hat \beta_1)\right)} \right]\sum_{x_i=1}y_i=n_1\cdot \left[\frac {\phi(\hat \beta_0+\hat \beta_1)}{\left(1-\Phi(\hat \beta_0+\hat \beta_1)\right)} \right]$$
$$\Rightarrow \Phi(\hat \beta_0+\hat \beta_1)=\frac 1{n_1}\sum_{x_i=1}y_i= \hat p_{1|1} \qquad [4]$$
In an analogous manner, working eq. $[3]$ we obtain
$$ \Phi(\hat \beta_0) = \frac 1{n_0}\sum_{x_i=0}y_i= \hat p_{1|0}\qquad [5]$$
So the MLE is calculated so as the LHS's are nothing more than the sample estimates of the conditional expected values of the dependent variable.  
We have now to "bypass" the fact that $n_1$ and $n_0$ are random variables themselves, and dependent for that matter, since $n_0 = n-n_1$ (we treat the sample size as deterministic). So first we consider the conditional covariance
$$\begin{align} \operatorname {Cov}\left(\Phi(\hat \beta_0+\hat \beta_1),\Phi(\hat \beta_0) \mid n_1,n_0\right) = E\left (\frac 1{n_1}\sum_{x_i=1}y_i\frac 1{n_0}\sum_{x_i=0}y_i\mid n_1,n_0\right) &\\- E\left (\frac 1{n_1}\sum_{x_i=1}y_i\mid n_1,n_0\right)E\left (\frac 1{n_0}\sum_{x_i=0}y_i\mid n_1,n_0\right)\end{align}$$
$$=\frac 1{n_1n_0}E\left[\sum_{i\neq j}y_iy_j\right] - \frac 1{n_1}E\left[\sum_{x_i=1}y_i\right]\frac 1{n_0}E\left[\sum_{x_i=0}y_i\right]$$
and using the i.i.d. assumption
$$=\frac 1{n_1n_0}n_1n_0p_{1|1}p_{1|0} - \frac 1{n_1n_0}n_1n_0p_{1|1}p_{1|0}=0$$
The conditional covariance is zero, and so its expected value will also be zero. Then, by the Law of Total Variance, we are left with
$$\operatorname {Cov}\left(\Phi(\hat \beta_0+\hat \beta_1),\Phi(\hat \beta_0) \right) =\operatorname {Cov}\left(E\left (\frac 1{n_1}\sum_{x_i=1}y_i\mid n_1,n_0\right),\;E\left (\frac 1{n_0}\sum_{x_i=0}y_i\mid n_1,n_0\right)\right)$$
$$=E(p_{1|1}p_{1|0}) - E(p_{1|1})E(p_{1|0})$$
But these are the theoretical probabilities, i.e. constants. So 
$$\operatorname {Cov}\left(\Phi(\hat \beta_0+\hat \beta_1),\Phi(\hat \beta_0) \right) =0$$
also. Is this intuitive or counter-intuitive?  
