Joint Probability of Sum Normal

[zli034]

Don't know there are anyone can help me out with this. This is just something I asking myself, not a homework I must say.

Let's define X and Y are 2 standard normal random variables. And random variable Z=X+Y.

For real number a, we know P(X>a), the probability of X is greater than the real number a.
For real number b, we know P(Y>b), the probability of Y is greater than the real number b.
For real number c, we know P(Z>c), the probability of Z is greater than the real number c.

These are simple things.

We also can determine a joint probability P(X>a and Y>b), the probability of X is greater than a, also Y is greater than b. Since X and Y are independent, this joint probability is still simple to know.

What about joint probability P(X>a and Y>b and Z>c)? Because Z is correlated with both X and Y, I don't know how to do this. Thanks for help

 

[wayneckm]

I think you can use the relationship $$P(X>a \text{ and } Y>b \text{ and }Z>c) = P(Z>c | X>a \text{ and }Y>b)P(X>a \text{ and }Y>b)$$

 

[zli034]

I think you can use the relationship $$P(X>a \text{ and } Y>b \text{ and }Z>c) = P(Z>c | X>a \text{ and }Y>b)P(X>a \text{ and }Y>b)$$

This relationship is certainly true. However I still don't get this part, P(Z>c| X>a and Y>b).
when a+b>c, then P(Z>c|X>a and Y>b)=1. And how about a+b<c? How do I express this step function? Or there is just no such function exist.

 

[EnumaElish]

Put y on the vertical axis, x on the horizontal.

y = b is a horizontal line.
x = a is a vertical line.
x+y = c is a downward-sloping line from (x,y) = (0,c) to (c,0).

As long as x > a and y > b, x+y > c is satisfied if c < a+b. For those values, Prob(x>a, y>b, z>c) = Prob(x>a, y>b).

If c > a+b, you need to figure in the additional constraint z>c. In this region {x>a, y>b, z>c given c > a+b}, Prob(x>a, y>b, z>c) is given by the double-integral of the joint pdf of X and Y, f(x,y) = f(x)f(y), first w/r/t x, from x = max(a, c - y) to infinity, then w/r/t y, from y = b to infinity.