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For $X \sim N(\mu_x,\sigma_x^2)$ and $Y\sim N(\mu_y,\sigma_y^2)$ which are independent, can I calculate $P(X+Y<0 | X<0)$?

I found this question for independent standard r.v and this one for correlated standard r.v, however I could not seem to understand how to translate the calculations for non-standard r.v.s like mine? Is there a way to do this?

Should I do it like the following?

Let $Z =X+Y$, then, $$ P(X+Y<0 | X<0) = \int_{-\infty}^0\Phi(z) \int_{-\infty}^0 \Phi(\frac{x-\mu_x}{\sigma_x}) dxdz \\ =\int_{-\infty}^0\Phi(z) \int_{-\infty}^{\frac{-\mu_x}{\sigma_x}} \Phi(x) dxdz $$

coffee-raid
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1 Answers1

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Observe that $X=\mu_x+\sigma_x Z$ and $Y=\mu_y+\sigma_y W$ where $Z,W$ are independent standard normal random variables. Thus, $$ \mathbb P(X+Y<0\mid X<0)=\mathbb P(\mu_x+\sigma_x Z+\mu_y+\sigma_y W<0\mid \mu_x+\sigma_x Z<0). $$ We can expand this as a ratio of probabilities where the denominator is given by $$ \mathbb P(\mu_x+\sigma_x Z<0)=\mathbb P\left(Z<\frac{-\mu_x}{\sigma_x}\right)=\int_{-\infty}^{-\mu_x/\sigma_x}\Phi(z)\ dz, $$ where $\Phi(z)$ denotes the density of the standard normal distribution. Similarly, the numerator is given by $$ \mathbb P(\mu_x+\sigma_x Z+\mu_y+\sigma_y W<0, \mu_x+\sigma_x Z<0)= \mathbb P\left(Z<\frac{-\mu_x}{\sigma_x}, W < \frac{-\mu_x-\mu_y-\sigma_x Z}{\sigma_y}\right). $$ To calculate the latter probability, we can start with the conditional probability $$ \mathbb P\left(W < \frac{-\mu_x-\mu_y-\sigma_x Z}{\sigma_y}\mid Z\right)= \int_{-\infty}^{(-\mu_x-\mu_y-\sigma_x Z)/\sigma_y}\Phi\left(w\right)\ dw. $$ Integrating yields the expression $$ \mathbb P\left(Z<\frac{-\mu_x}{\sigma_x}, W < \frac{-\mu_x-\mu_y-\sigma_x Z}{\sigma_y}\right)=\int_{-\infty}^{-\mu_x/\sigma_x}\Phi(z)\int_{-\infty}^{(-\mu_x-\mu_y-\sigma_x z)/\sigma_y}\Phi\left(w\right)\ dw\ dz. $$

pre-kidney
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