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In section 11.3.1 of Introduction to probability models by Ross (10th edition), a very strange phenomenon is described. If you take two independent standard normal distributions and sum their squares, you get an exponential distribution with rate $\frac{1}{2}$. This is proven mechanically. I'm looking into some intuitive insight into this. The exponential distribution is known to be memory-less. And it seems this sum of squares of two i.i.d. Gaussians is also memory-less. For the exponential distribution ($X$), this means (from some $t>s$):

$$P(s <X\; \& \; X<t) = P(X<t | s<X)P(s<X) = P(X>s)P(X<t-s)$$

Now for two Gaussians, $Y$ and $Z$ we get:

$$P(Y^2+Z^2 < r^2+t | Y^2+Z^2>r^2) = P(Y^2+Z^2<t)$$

Is there some connection to a fundamental property of the Gaussian that helps get an intuitive explanation for this memoryless behavior?

Rohit Pandey
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  • What is a memoryless distribution ? I thought memoryless ( Markov ) property was just a property of stochastic processess... can you give some reference ? – Thomas Jan 23 '22 at 14:35
  • @Thomas - the exponential distribution is like tossing a coin every micro-second and the event happens whenever you get a heads (which is vanishingly rare). Hence memory-less because the coin tosses are independent. It is the continuous limit of the geometric distribution. See chapter 5 of introduction to probability models by Ross. Will link relevant section a little later. – Rohit Pandey Jan 23 '22 at 18:45
  • And indeed, continuous time Markov chains have the exponential distribution fundamentally embedded within them (chapter 6 of same book). – Rohit Pandey Jan 23 '22 at 18:50
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    From here and here, I feel like this may not have a very intuitive explanation : but the problem is that we don't know there isn't one! I tried to have a couple of stabs at it, but these are still the best I could find. I'm sorry for the discouragement, but I'm following this question in any case, and +1. – Sarvesh Ravichandran Iyer Jan 24 '22 at 18:35

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