I am trying to find $P(z \le -2)$ where Z~N(0,1).
I have done it using tables, but for a challenge I am trying to do it using integration.
I know how to integrate the standard normal distribution for infinite limits, in order to show that the total probability is 1. However, I am getting stuck when it comes to integrating over the interval $(-\infty , -2]$.
I have the following:
$$I^2=\frac{1}{2 \pi} \int_{-\infty}^{-2} \int_{-\infty}^{-2} e^{-\frac{1}{2}(x^2+y^2)} dy dx $$
Now when I change to polar coordinates, obviously the integrand becomes $e^{\frac{-r^2}{2}}$ but what do the upper limits change to?
I am not that clever so if this is a major thing, please just tell me that, I won't be offended.
Thanks in advance!
The problem is that when you change the limits of integration, you'll just get something else you can't compute in closed form. The region of integration is an infinite rectangle in the fourth quadrant, so we have $-{\pi\over2}<\theta<-\pi.$ To figure out the range of $r$ we need to split into the two cases $-{\pi\over2}<\theta<{-3\pi\over4}$ and ${-3\pi\over4}<\theta<-\pi.$
Let's look at the second case. We need to figure out where the line $y=x\tan{\theta}$ intersects the line $x=-2,$ which is clearly at $(-2, -2\tan{\theta})$ so that the minimum value of $r$ is $2\sqrt{1+\tan^2{\theta}}=2|\sec{\theta}|=-2\sec{\theta},$ since $\theta$ is in the third quadrant. Of course, the maximum value of $r$ will be $\infty$. Now you can set up the integral, but after you integrate out $r$, you'll be left with $\int e^{-2\sec^2{\theta}}d\theta,$ which is not elementary.