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I am studying Monte Carlo simulation and I came across with this claim from notes:

Let $Y_1$ and $Y_2$ be two independent $Exp(1)$ random variables. We accept $Y_1$ as one sample when $Y_2$ is larger than $(Y_1 - 1)^2/2$. Then by the memory-less property of exponential random variables, $Y_2 - (Y_1 - 1)^2/2$ is also $Exp(1)$ distributed and independent of $Y_1$. This means we can make use of $Y_2 - (Y_1 - 1)^2/2$ as a new exponential r.v. in the next iteration.

I am kind of stuck with how to apply the memory-less property here. Any solution or hint will be appreciated. Thanks!

Van Tom
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    Given that $Y_2-(y_1-1)^2/2$ can take negative values, the property cannot hold. – Xi'an Jan 31 '21 at 12:28
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    @Xi'an I mean in the case when $Y_2$ exceeds $(Y_1-1)^2/2$. I have edit the original post. Thanks for pointing this out! – Van Tom Jan 31 '21 at 13:20
  • Hint: neither the distribution $Y_1$ nor the function $f: y\to (y-1)^2/2$ affect the conclusion that $Y_2-f(Y_1)$ has an Exponential distribution, provided the values of $f(Y_1)$ are almost surely non-negative. Contemplating a scatterplot of $(f(Y_1),Y_2-f(Y_1))$ might help. – whuber Jan 31 '21 at 15:59

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