2
$\begingroup$

$\newcommand{\Exp}[1]{\mathbb{E}\left[#1\right]}$ I am interested in understanding wether the following approach holds when calculating the expectation of two correlated random variables. Suppose $X\sim F_X$ and $Y\sim F_Y$ have positive support, and assume the joint density may be implied from the standard Gaussian copula with correlation $\rho$. It then follows

$$ \Exp{\min\{X,Y\}}=\iint_{\mathbb{R}^2_{+}}\min\{x,y\}c_\text{Gauss}(F_X(x),F_Y(y))f_X(x)f_Y(y)d(x,y)$$ where $c_\text{Gauss}$ denotes the density of the copula and $f_X,f_Y$ are the p.d.f. of $X$ and $Y$ respectively. Set $u=F_X(x)$ and $v=F_Y(y)$ yielding

$$\iint_{[0,1]\times[0,1]}\min\{F^{-1}(u),F^{-1}(v)\}c_\text{Gauss}(u,v)d(u,v)$$ and expanding the min function by cases and using the fact that c.d.f. are continuous/monotonic/increasing, we have \begin{align*} \Exp{\min\{X,Y\}}&=\iint_{[0,1]\times[0,1]}\min\{F_X^{-1}(u),F_Y^{-1}(v)\}c_\text{Gauss}(u,v)d(u,v)\\ &=\int_0^1\int_0^uF_X^{-1}(u)c_\text{Gauss}(u,v)dvdu+\int_0^1\int_0^vF_Y^{-1}(v)c_\text{Gauss}(u,v)dudv \end{align*} and since the integrand is non-negative and measurable, we may apply Tonelli, \begin{align*} \Exp{\min\{X,Y\}}&=\int_0^1F_X^{-1}(u)\left(\int_0^uc_\text{Gauss}(u,v)dv\right)du+\int_0^1F_Y^{-1}(v)\left(\int_0^vc_\text{Gauss}(u,v)du\right)dv\\&= \int_0^1F_X^{-1}(u)\frac{\partial C_\text{Gauss}(u,u)}{\partial u}du+\int_0^1F^{-1}_Y(v)\frac{\partial C_\text{Gauss}(v,v)}{\partial v}dv \end{align*} where $C_\text{Gauss}$ denotes the bivariate standard normal copula. Is this correct? And if it is, can other subtle simplifications be made?

$\endgroup$
1
  • $\begingroup$ Looks good to me. $\endgroup$
    – zhoraster
    May 29 at 16:02

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.