# X and Y are two independent standard gaussian variables, what is the Probability that X>100*Y

probabilityrandom variablesstatistics

We want to find $$P(X> 100*Y)$$.

My strategy was to first express $$X>100*Y$$ as $$X-100*Y>0$$ and express $$Z=X-100*Y$$.
$$Z$$ should be another normal random variable, yet not standard. My approach further was to standardise it so I can use 68–95–99.7 rule. However this approach is leading nowhere.

Can anyone give any hints how I should approach it?

I really like your method of using $$Z = X - 100\cdot Y$$

You were definitely correct in pointing out $$Z$$ is again Normal! We can characterize every normal distribution by its mean and variance. So to find out $$Z$$ exactly we just need these two things.

Mean of Z

You will have seen before that expectation is "linear" that is we know $$\mathbb{E}[Z] = \mathbb{E}[X] - 100\cdot \mathbb{E}[Y] = 0 - 100\cdot 0 = 0$$

Variance of Z

This is slightly harder, in general variance doesn't add up nicely but because $$X$$ and $$Y$$ are independent then it will!
Var$$[Z] =$$ Var$$[X] + 100^2$$Var$$[Y] = 1+10,000 = 10,001$$

Finishing off

Hence $$Z \sim N(0,10001)$$
And because Normal distributions are symmetric about their mean we know $$\mathbb{P}[Z > 0 ] = \frac{1}{2}$$

(Yes we didn't need to calculate the variance at all)

Another approach

I first saw this question a while back and used a different method. Yours I think is much better but I will include mine here.

We find the probability of $$X>100\cdot Y$$ by conditioning on $$4$$ events, namely the signs of the two random variables. Notice that as $$X$$ and $$Y$$ are standard normal they both have equal probability of being positive or negative.

1. $$X>0$$ , $$Y<0$$. In this case we know for certain $$X>100Y$$
2. $$X<0$$, $$Y > 0$$. In this case we know for certain $$X < 100Y$$
3. $$X>0$$, $$Y>0$$. Let $$p$$ be the probability of $$X>100Y$$ when they are both positive.
4. $$X<0$$ , $$Y<0$$. By symmetry the probability of $$X>100Y$$ is $$1-p$$

Hence $$\mathbb{P}[X>100Y] = \frac{1}{4}(1+0+p+1-p) = \frac{1}{2}$$