probability calculation with two independent normal RVs

Question

I don't understand why

Probability[X < Y, {X \[Distributed] NormalDistribution[0, 1],
  Y \[Distributed] NormalDistribution[0, 1]}]

gives 1/2

while we know that X/Y is Cauchy(0,1) and that

Probability[Z < 1, Z \[Distributed] CauchyDistribution[0, 1]]

gives 3/4

I tried many different approaches and I can't figure out why we have this difference.

Answer 1

Question

If $X \sim N(0,1)$ and $Y \sim N(0,1)$ are independent, how come:

$$ P(X<Y) = \frac12 \quad \text{while} \quad P(\frac{X}{Y}<1) = \frac34$$

Answer

Nothing to do with Mma. The answer is that the rules of standard algebra do NOT apply to the algebra of random variables, so you cannot simply divide both parts of $P(X<Y)$ by $Y$ ... to get $P(\frac{X}{Y}<1)$ ... , which is what you seem to be doing. Note that $\frac{Y}{Y} \neq 1$, since this denotes the ratio of two random variables (not 2 fixed numbers).

Similarly: $P(X<0) \neq P(X^2<0) \quad$ (The former is $\frac12$; the latter is 0)

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P.S. An easy solution to $P(X<Y)$ is:

• Let $Z = (X-Y) \sim N(0, 2)$.

• $P(X<Y) = P(X-Y<0) = P(Z<0) = \frac12$ .