Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have:

$ V_1 = X_1+A $ where $A>0$ is some constant $ V_2 = X_2+B $ where $B>0$ is some constant

Furthermore assume that $X_1$ and $X_2$ are independent and distributed with a standard normal distribution $N(0,1)$.

I want to get $P(V_1+V_2 > 1\ \&\ V_1,V_2 \in [0,1] ) $

Here's what I'm thinking:

$V_1+V_2 > 1 \implies X_1+X_2>1-A-B \implies X_1>1-A-B-X_2$ $V_2 \in [0,1] \implies -A<X_2<1-A $

so the answer would be:

Then, I could do a double integral one from $1-A-B-X_2$ to $\infty$ for $X_1$ and the other from $-A$ to $1-A$ for X_2, but I'm still not sure what I'm integrating over, e.g pdf of the sum?

share|cite|improve this question
up vote 1 down vote accepted

Not quite.

You want $0 \le V_1 \le 1$ but also $1-V_2 \lt V_1$.

Since you have $0 \le V_2 \le 1$, you can turn your constraints on $V_1$ into $1-V_2 \lt V_1 \le 1$.

Since $V_1 = X_1+a$ and $V_2 = X_2+b$ (it is conventional to use lower case for constants), these become $$-b \le X_2 \le 1-b$$ $$1-a-b-X_2 \lt X_1 \le 1-a$$ and so your integral becomes $$\int_{x_2=-b}^{1-b} \int_{x_1=1-a-b-x_2}^{1-a} \phi(x_1)\phi(x_2) \, dx_1 \, dx_2$$ where $\phi(x)$ is the probability density function of a standard normal distribution.

share|cite|improve this answer
Thanks for the help. I'm wondering if that integral can be simplified. I end up with an integral $ \int_{-b}^{1-b} \Phi(1-a-b-x_2) \phi(x_2)dx_2 $ Can this be simplified? Note, I have other terms, but was able to simplify them. – Greg Apr 22 '12 at 2:20
You should have got $\int_{-b}^{1-b} (\Phi(1-a) - \Phi(1-a-b-x_2)) \phi(x_2)dx_2$ or $\Phi(1-a)(\Phi(1-b)-\Phi(-b))+ \int_{-b}^{1-b} \Phi(1-a-b-x_2) \phi(x_2)dx_2$. Perhaps you did. I doubt there is a further simplification. – Henry Apr 22 '12 at 10:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.