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Suppose we have two independent standard normal random variables, $Z_1, Z_2$, say. I wanted to calculate this probability

$$\mathbb{P}(Z_1>0,Z_1+Z_2<0).$$

I can kind of explain the probability via a diagram—I plotted $Z_2$ against $Z_1$ and said the probability must be a $\frac{1}{8}$ but I wanted a more rigorous approach. Could someone please guide me through how to approach this probability in a more rigorous manner, in particular, if you could explain where you have used independence-that would be great!

Xander Henderson
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UnsinkableSam
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    Two i.i.d. normal random variables give a bivariate normal distribution with circular symmetry about the mean (using independence and identical) which for standard normals is a mean of $(0,0)$. The circular symmetry results in your diagram being rigorous. – Henry Sep 20 '20 at 18:52
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    You can just count the number of cases where $Z_1>0$ and $Z_1+Z_2<0$ as shown in this answer. You can also answer this by using this result. – StubbornAtom Sep 20 '20 at 18:53

2 Answers2

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We have \begin{align*} P(A > 0, A+B < 0) &= P(A > 0, B < 0, A+B < 0) \\ &= P(A > 0, B < 0) \cdot P(A < -B \mid A>0, B<0) \\ &= P(A > 0, B < 0) \cdot P(|A| < |B| \mid A>0, B<0) \\ &= P(A > 0, B < 0) \cdot P(|A| < |B|) \\ &= \frac{1}{4} \cdot \frac{1}{2} \end{align*} The last three inequalities use the fact that they are independent normal with mean 1.

eatfood
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  • Thank you for your brilliant answer! Just one small detail, why is it true that $P(|A|<|B|)=\frac{1}{2}$? – UnsinkableSam Sep 20 '20 at 18:38
  • Its by symmetry: If two (continuous) variables are identically distributed, then the probably of the first being larger than the second is $1/2$. – eatfood Sep 20 '20 at 19:15
  • More precisely if $A$ and $B$ are iid, then you have $P(A < B) + P(B < A) + P(A = B) = 1$. Now, by iid, we can interchange A and B of the 2nd term of the LHS, so $P(A < B) + P(A < B) + P(A = B) = 1$, and thus $2P(A < B) = 1 - P(A=B)$, then the rest is easy. – eatfood Sep 20 '20 at 19:20
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This is essentially the same as eatfood's answer, written with some more detail. I've tried to pinpoint exactly which parts of the hypotheses we use and where. Clearly, the statement is true in greater generality.

First, let $Z_3=-Z_2$. Then $Z_3\sim N(0,1)$ and $Z_1, Z_3$ are independent. Then we want $Z_1>0,\ Z_3>Z_1$. $$ \begin{align} \Pr(Z_1>0, Z_3>Z_1)&=\Pr(Z_1>0)\Pr(Z_3>Z_1\mid Z_1>0)\\ &=\frac12\Pr(Z_3>Z_1\mid Z_1>0)\\ &=\frac12\left(\frac12\Pr(Z_3>Z_1\mid Z_1>0,Z_3>0)+\frac12\Pr(Z_3>Z_1\mid Z_1>0,Z_3\leq0)\right)\\ &=\frac14\Pr(Z_3>Z_1\mid Z_1>0,Z_3>0)\\ &=\frac14\cdot\frac12=\frac18 \end{align}$$

It's only in the last line that we use the fact that $Z_1, Z_3$ are i.i.d. Up to there, the only thing we've used specific to the problem at hand is that the the probability that a $N(0,1)$ r.v. takes a positive value is $\frac12$.

To get to the last line, we have by symmetry, $$\Pr(Z_3>Z_1\mid Z_1>0,Z_3>0)=\frac12(1-\Pr(Z_1=Z_3|Z_1>0,Z_3>0))=\frac12$$ since $Z_1, Z_3$ are continuous random variables without atomic points.

I think the last condition is enough, though it may not be the precise statement I want. Anyway, it's obviously true for $N(0,1)$ r.v.s: $$\Pr(Z_1=Z_3|Z_1>0,Z_3>0)=\frac4{2\pi}\int_0^\infty e^{-x^2}\int_x^xe^{-y^2}\,\mathrm{d}y\,\mathrm{d}x=0$$

saulspatz
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