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I'm using the function $$\frac{xe^{-x^2}}{\int_0^1 te^{-t^2} \, dt}$$ as a probability density function for generating numbers between 0 and 1. If I want to increase the probability to the zero (crush it to the zero), I have to increase the exponent using something like $xe^{-8x^2}$, but how can I do this if I want to increase the probability value of the function to zero (to crush it to zero in a way to have a pick to the zero)?

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put on hold as unclear what you're asking by Jonas Meyer, Rolf Hoyer, pjs36, Chappers, Joe Johnson 126 14 hours ago

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I edited this for readability and grammar, but I have no idea what is being asked. –  Matthew Conroy May 22 '11 at 20:59
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If pick to the zero means peak at zero, simply get rid of the factor $x$ and keep the exponential. –  Did May 31 '11 at 22:56

1 Answer 1

Well, one thing you could do is generate you variable, $X$, from a mixture density

$$ p(x) = \lambda p_{1}(x) + (1-\lambda) p_{2}(x) $$

which is a density as long as $\lambda \in (0,1)$ and $p_{1},p_{2}$ are densities. You could choose $p_{1}$ to be the density you specified above and $p_{2}$ to be, for example, a point mass at 0 (e.g. $p_{2}(0) = 1$ and $p_{2}(x) = 0$ for $x \neq 0$). Specifically, the algorithm to sample from $p$ would be:

  1. Generate a bernoulli trial, $Z$, with success probability $\lambda$
  2. If $Z=1$ then generate from $X$ from $p_{1}(x)$. If $Z=0$, then $X=0$.

Of course, a better approach would probably be to use the Beta distribution which is a broad distribution supported on $(0,1)$. Parameter values can be chosen to achieve an arbitrary amount of mass near 0.

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