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?

## Best Answer

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 ZYou 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 ZThis 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 offHence $Z \sim N(0,10001)$

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 approachI 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.

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