# X, Y are iid from N(0,1). What's the probability that X>2Y

I was thinking, since $X, Y$ are from $N(0,1)$ and they are independent, then

$X - 2Y$ has a distribution of $N(0, 5)$. Then $X-2Y > 0$ has probability of $1/2$.

The above seems correct to me, though it seems like then $X>nY$ would have probability of $1/2$. That seems a bit wrong. Did I get anything wrong?

• What seems 'a bit wrong' there? Are you thinking about the conditional probability perhaps? ($P(X>nY|Y)$ ... that's not the probability in question) Apr 5, 2015 at 23:43
• If I understood you right the results $\frac{1}{2}$ seems not intuitive for you. But even in case if n is large Y is positive with probability $\frac{1}{2}$ (and negative with probability $\frac{1}{2}$). Although |X| is unlikely to be larger than |nY|, the probability without absolute values is reasonalby $\frac{1}{2}$.
– Lan
Apr 7, 2015 at 20:56

With a bivariate standard normal (i.e. iid standard normal), the probability of lying on one side of a line through the origin is $\frac{_1}{^2}$ no matter what the slope of the line is. This follows, for example, from the rotational symmetry of the bivariate distribution about $O$, since we could rotate the problem to one of considering $P(X'\gt0)$ in rotated coordinates.
Indeed, considering the use of affine transformations means it must be $\frac{_1}{^2}$ much more generally -- the argument will apply to any bivariate normal where both variances are greater than 0.
• If $X$ and $Y$ are zero-mean jointly normal random variables (not necessarily independent), then $X-aY$ is a zero-mean normal random variable and thus $$P\{X > aY\} = P\{X-aY > 0\} = \frac 12.$$ Independence and variances have nothing to do with the matter: all that is needed for the above result to hold is that the variables be jointly normal and that the means be zero. (Trivial exception when $X-aY$ equals $0$, that is, it is a degenerate normal random variable a.k.a. a constant which happens when $X$ and $Y$ are perfectly correlated and $\sigma_X = a\sigma_Y$). Apr 6, 2015 at 2:48