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this is what chat GPT gave me, wanted to make sure it's correct and why?

The variables X and Y are both standard normal random variables, i.e., they follow a normal distribution with mean 0 and standard deviation 1.

The variable Z = X + Y is also a normal random variable. The mean of Z is the sum of the means of X and Y, which is 0 + 0 = 0. The variance of Z is the sum of the variances of X and Y (since X and Y are independent), which is 1 + 1 = 2. Therefore, Z is a normal random variable with mean 0 and standard deviation $$\sqrt{2}$$.

The event that both X and Z are positive corresponds to the upper right quadrant of the (X, Z) plane. Because X and Z are bivariate normally distributed, this region corresponds to a quarter of the total probability, so the probability that both X and Z are positive is 0.25 or 25%.

This is because the joint distribution of (X, Z) is symmetric about both the X-axis and the Z-axis, and the four quadrants of the (X, Z) plane are equivalent.

So, the probability that both X and X+Y are positive is 0.25 or 25%.

Armaghan Naveed
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    you need the joint distribution to answer the question... – Surb Dec 24 '23 at 12:11
  • OK, you used ChatGPT, but what do you understand about what ChatGPT said, and what parts are you simply regurgitating? Also I find it weird you used Marhjax to format your math expressions in only one spot and not the others. – Divide1918 Dec 24 '23 at 12:12
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    $X$ and $Y$ normal does not necessarily make $X+Y$ normal. ChatGPT is not for Mathematics. – geetha290krm Dec 24 '23 at 12:19
  • Note: the OP does in fact declare that $X,Y$ are independent, though, confusingly, that's just tossed in via a parenthetic clause. For clarity, that assumption should have been made explicitly at the start. – lulu Dec 24 '23 at 12:50
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    I’m voting to close this question because OP used chatGPT. – Kurt G. Dec 24 '23 at 12:56
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    @KurtG. I disagree with your reaction. I am not aware of any rules that prohibit the OP (i.e. original poster) from attempting to use analysis from chatGPT, rather than a much more reliable source, such as a math textbook. In judging whether a question should be closed, to me, the issue is not whether the OP has gone off the rails. Instead, the issue is whether the OP has made a significant attempt (of some sort) to tackle the problem, and has made a reasonable attempt to document his work directly in the posted question. In your opinion, does this posting meet those standards? – user2661923 Dec 24 '23 at 14:59
  • To the OP (i.e. original poster): Have you ever taken a college course in statistics? Personally, I have not, which is why, if I was invited to attack this problem, I would pass. That is, since I have no (formal, well organized) education in this part of math, I believe that I would be in over my head. – user2661923 Dec 24 '23 at 15:03

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That ChatGPT answer is completely incorrect.

First, it looks like ChatGPT has assumed that $X$ and $Y$ are independent, which wasn't part of the original prompt (assuming the question title is the verbatim prompt). This extra assumption isn't made clear, except as a parenthetical note in only one, non-critical spot in the argument. Other, more critical parts of the argument -- $Z$ is normally distributed, the joint distribution of $X$ and $Z$ is bivariate normal -- rely on the independence assumption, but ChatGPT doesn't explain this.

Second, and more critically, even with the independence assumption added, ChatGPT incorrectly concludes that the bivariate normal distribution of $X$ and $Z$ is symmetric about the $X$ and $Z$ axes. This is false. So, the final conclusion that the probability is 0.25 is also false.

It turns out that the correct answer under the assumption that $X$ and $Y$ are independent is 3/8. This follows from a "radial symmetry" argument as in this answer. Note that the probability given in that answer (of 3/4) is the conditional probability of $X+Y>0$ given $X>0$, but to calculate this, the answer identifies 8 equally likely cases where exactly 3 have both $X>0$ and $X+Y>0$, giving the probability 3/8.

K. A. Buhr
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