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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?

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  • $\begingroup$ 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) $\endgroup$
    – Glen_b
    Apr 5, 2015 at 23:43
  • $\begingroup$ 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 $\frac{1}{2}$. $\endgroup$
    – Lan
    Apr 7, 2015 at 20:56

1 Answer 1

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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.

enter image description here

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.

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    $\begingroup$ Thanks, I just found my conclusion a bit counterintuitive, but your diagram makes all this clear to me. $\endgroup$ Apr 6, 2015 at 0:12
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    $\begingroup$ 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 $\sigma_X = a\sigma_Y$). $\endgroup$ Apr 6, 2015 at 2:48
  • $\begingroup$ Thanks Dilip, your comment is of course completely correct - I was starting with the conditions as given and attempting to give some motivation for the result the OP had already derived. $\endgroup$
    – Glen_b
    Apr 6, 2015 at 2:54

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