If I have two jointly Gaussian RVs $X,Y$, How can I express $P(X+Y>0)$ in terms of the Q-function?
If linear combinations of jointly gaussian RVs are also jointly gaussian, so if $Z=X+Y$, then is $P(X+Y>0)$ = $Q(Z)$ evaluated at 0?
Q-function of jointly Gaussian RVs
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probability-distributions
normal-distribution
1 Answers
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If $X,Y$ are bivariate normal then their sum $Z=X+Y$ is normal with mean $$\mu_z = E(X) + E(Y) = \mu_x+\mu_y $$ and variance $$\sigma^2_z = \mathrm{Var}(X) +\mathrm{Var}(Y) +2\mathrm{Cov}(X,Y) = \sigma_x^2 +\sigma_y^2 + 2\sigma_x\sigma_y\rho_{xy} $$ so $$Z'= (Z-\mu_z)/\sigma_z$$ is a standard $N(0,1).$
Thus, we have $$ P(X+Y>0) = P(Z>0) = P(Z'\sigma_z+\mu_z >0) = P(Z'>-\mu_z/\sigma_z) = Q(-\mu_z/\sigma_z).$$