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Given IID $X_1, X_2 \sim \mathcal{N}(0,1)$, we want to determine $P(X_1 > 0 \mid X_1 + X_2 > 0)$.

This is what I think is the approach for this problem:

\begin{align} P(X_1>0\mid X_1 + X_2 > 0) = P(X_1 > 0 \mid X_1 > -X_2) \\ P(X_1 \leq x \mid X_1 > (-X_2=x_2)) = \int_{-x_2}^x \frac{1}{\sqrt{2\pi}}\exp(-\frac{1}{2}x_1^2) \, dx_1 \end{align}

I integrate the above to get $P(X_1 \leq x \mid X_1 > (-X_2=x_2))$, and then I integrate $P(X_1 \leq x_1\mid X_1 > (-X_2=x_2))$ over all $x_2$: $$ P(X_1 \leq x_1\mid X_1 > X_2) = \int_{-\infty}^\infty P(X_1 \leq x_1\mid X_1 > (-X_2=x_2)) \, dx_2 $$

Is this a correct approach? I feel that there is something simpler than this.

Michael Hardy
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user5965026
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  • You write $P(X_1 \leq x_1|X_1 > (-X_2=x_2)) = \int_{-x_2}^{\infty} \frac{1}{\sqrt{2\pi}}\exp(-\frac{1}{2}x_1^2) dx_1$ which doesn't really make sense. On the LHS, $x_1$ is the integration variable, and doesn't factor into the LHS at all. – Varun Vejalla Aug 06 '20 at 04:12
  • Oh right. $x_1$ on the LHS should just be $x$. – user5965026 Aug 06 '20 at 04:14
  • @VarunVejalla Also had a typo in upper limit of integral. Should be $x$ instead of $\infty$ – user5965026 Aug 06 '20 at 04:15
  • Also, what you have there is $P(X_1 \le x \cap X_1 > (-X_2 = x_2))$, not $P(X_1 \le x | X_1 > (-X_2 = x_2))$. You would need to divide by $P(X_1 > (-X_2 = x_2))$ on the RHS to make it correct. – Varun Vejalla Aug 06 '20 at 04:18
  • @VarunVejalla I was hesitant about my expression. But I don't think I see why it's the former, and not the latter. Could you explain that part? We are given that $X_1 > X_2$, so shouldn't the lower limit of $X_1$ be $X_2 = x_2$? – user5965026 Aug 06 '20 at 04:20
  • It is the formula for conditional probability. Specifically, your current expression doesn't account for the fact that $X_1 > (-X_2 = x_2)$ is given information, so you would need to divide by the probability of that happening. – Varun Vejalla Aug 06 '20 at 04:21
  • @VarunVejalla Ah, the issue here is I was integrating over a marginal pdf of $X_1$ and not the conditional pdf of $X_1$, right? – user5965026 Aug 06 '20 at 04:22
  • I'm not very knowledgeable with probability distributions, but I think that's right. – Varun Vejalla Aug 06 '20 at 04:29

3 Answers3

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The formula for conditional probability is $$P(A | B) = \frac{P(A \cap B)}{P(B)}$$

This then means that the answer is $$\frac{P(X_1 > 0 \cap X_1 + X_2 > 0)}{P(X_1 + X_2 > 0)}$$

Clearly, the denominator is $1/2$. Then for the numerator, the integral is $$\iint_{R} p(x_1) p(x_2) \,dx_1 \,dx_2$$

where $R$ is the region satisfying $x_1>0, x_1 + x_2 > 0$ and $p(x)$ is the equation for the probability distribution.

Plugging everything in, the integral comes out to $\frac{3}{8}$, which means the final probability comes out to $$\frac{\frac{3}{8}}{\frac{1}{2}} = \frac{3}{4}$$

Michael Hardy
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Varun Vejalla
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  • Is the denominator "clearly" 1/2 because $X_1 + X_2$ is normally distributed with mean 0? Is there another intuitive way to see why it is 1/2? I'm thinking about what you wrote for the numerator right now. – user5965026 Aug 06 '20 at 04:23
  • Yeah, that's right for the denominator. Another somewhat intuitive way is to pick positive $x, y$. Then out of $(x, y), (x, -y), (-x, y), (-x, -y)$, two will have positive sums and two will have negative sums. – Varun Vejalla Aug 06 '20 at 04:27
  • For the numerator, I think I am confused how you obtained $p(x_1) p(x_2)$. Is this product of pdfs the pdf for the probability in the numerator? – user5965026 Aug 06 '20 at 04:29
  • Yes, that's correct. The integral would be the probability of picking $x_1, x_2$ that falls in the region $R$. – Varun Vejalla Aug 06 '20 at 04:32
  • Makes sense. I'm really bad at forming integrals when it comes probability. In this case, I was thinking you would need to integrate the probability distribution of $X_1 + X_2$ instead – user5965026 Aug 06 '20 at 04:33
  • In Cartesian coordinates, how do you get the lower bound for $x_1$ in the region $R$? The upper bound should be $\infty$, but the lower bound is $\max(0, -Y)$ I believe. Seems the integral is hard to compute if you were to do this in cartesian instead of polar. – user5965026 Aug 06 '20 at 18:36
  • In my answer, I was just showing in very general terms, not going through the calculation of the integral. Doing it in polar would in fact be easier. With Cartesian coordinates, the bounds would be $x_1 > 0, x_2 > -x_1$. Another way to realize that the integral is $3/8$ is realizing that $3$ out of the $8$ sectors made by $x_1 = 0, x_2 = 0, x_2 = -x_1, x_2 = x_1$ are in the integral. – Varun Vejalla Aug 06 '20 at 18:44
  • Would the bounds in Cartesian look something like $\int_{-\infty}^{\infty}\int_{\max(0, -x_2)}^{\infty} \cdot dx_1dx_2$? This seems to satisfy $x_1 > 0, x_1+x_2 > 0$. – user5965026 Aug 06 '20 at 18:48
  • That would work, but simpler bounds are obtained by switching the order of integration: $\int_0^{\infty} \int_{-x_1}^{\infty} dx_2 dx_1$. – Varun Vejalla Aug 06 '20 at 18:49
  • Right good point. With my bounds, it seems you'd need to do some case work and compute the inner integral for the case where $0 > -x_2$ and for the case where $-x_2 > 0$. Sum the two cases and then integrate over $x_2$, so seems like it'd be more work than your formulation. – user5965026 Aug 06 '20 at 18:53
2

Is this a correct approach? I feel that there is something simpler than this.

There is no integration required when you use a graphical approach

When 3D-plotting the joint probability density against the $X_1,X_2$ plane, you obtain a radially-symmetrical bell-shaped 'pie' stretching infinitely but whose total volume measure is $1$.   So you can readily evaluate the probability of slices of this 'pie'.

$$\begin{align}\mathsf P(X_1>0\mid X_1>-X_2)&=\dfrac{\mathsf P(X_1>\max\{0,-X_2\})}{\mathsf P(X_1>-X_2)}\\[1ex]&=\dfrac{\mathsf P(X_1>-X_2>0)+\mathsf P(X_1>0\geq -X_2)}{\mathsf P(X_1>-X_2)}\\&~~\vdots\end{align}$$

Graham Kemp
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  • On the second line, it seems you are summing up the probability of disjoint events, right? essentially you're consider 2 disjoint events $\max{0,-X_2} = 0$ and $\max{0,-X_2} = -X_2$? – user5965026 Aug 06 '20 at 04:57
  • Yes, the partitioning is over $-X_2>0$ and $-X_2\leq 0$. – Graham Kemp Aug 06 '20 at 05:14
  • Got it. How do you graphically obtain numerical values for those probabilities? – user5965026 Aug 06 '20 at 05:17
  • Look to the relevant areas on a Cartesian plot. $\mathsf P(X_1<-X_2) = 1/2$ because $x<-y$ occupies half of the plane, and so forth. – Graham Kemp Aug 06 '20 at 06:25
  • Ah I figured it out graphically. Compared to Michael Hardy's approach above, which is essentially like yours but in polar coordinates, it seems yours may be more easily adaptable to the more general case where $X_1 > c$ and $X_1 + X_2 > d$ from constants $c,d$? – user5965026 Aug 06 '20 at 20:16
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In polar coordinates, divide the $(x_1,x_2)$-plane into eight sectors: $0 \le \theta < \pi/4,$ then $\pi/4\le \theta < 2\pi/4, $ and so on, up to $7\pi/4\le\theta < 8\pi/4.$

All eight of these have equal probability of being where the pair $(X_1,X_2)$ is, by symmetry.

Call the random angle $\Theta,$ so that $(X_1,X_2) = (\cos\Theta,\sin\Theta)\cdot \sqrt{X_1^2 + X_2^2}.$

Then the event $X_1+X_2>0$ is the same as $-\pi/4<\Theta< 3\pi/4$ (here we treat $-\pi/4<\Theta<0$ as being the same thing as $7\pi/4<\Theta< 0\text{ or } 0\le\Theta<3\pi/4,$ i.e. in effect we're adding modulo $2\pi.$

And the event $X_1>0$ is the same as $-\pi/2<\Theta< \pi/2.$

In effect $\Pr( X_1 > 0 \mid X_1+X_2>0)$ is the probability that $\Theta$ is in sector $7,8,1, \text{ or }2$ given that $\Theta$ is in sector $8,1,2,\text{ or } 3.$ And all eight sectors are equally probable.

So it's $3/4.$

Michael Hardy
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  • Yes, $3/4.$ It doesn't contradict that other one except in a typo in the last line, which I have now fixed. – Michael Hardy Aug 06 '20 at 05:16
  • I wasn't able to edit it because it didn't meet the character req. But the -pi/3 should be -pi/4. – user5965026 Aug 06 '20 at 05:43
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    @user5965026 : Fixed. For a thing like that you can go to the bottom of the posting and enter ${{}}$ and then you have enough characters. – Michael Hardy Aug 06 '20 at 16:16
  • Is this method adaptable to cases where $X_1 > c$ and $X_1 + X_2 > d$ for some constants $d,c$? It seems it would not be easy to adapt because the argument we made for the range of $\theta$ was simple to make because we could just divide the sample space into positives and negatives. But if instead instead our threshold is $c$ and $d$, then the argument no longer applies. – user5965026 Aug 06 '20 at 19:37
  • @user5965026 : The joint distribution of $(X_1,X_2)$ is symmetric about the origin and invariant under rotations centered at the origin, and that fact was relied upon heavily. So if you look at lines not passing through the origin, i.e. $c\ne0$ or $d\ne0,$ then it's nowhere near so simple. – Michael Hardy Aug 06 '20 at 22:12