Question on proving the inverse of the distribution function of the random variable
$begingroup$
We can approximate the conditional distribution by a random variable with density
$$
\pdelta_0(x)+(1-p)beta e^{-beta{x}}1_{x>0}dx
$$
that is to say a weighted mean between a Dirac mass at 0 and an exponential random variable with parameter $beta$. The constants $p in (0,1)$ and $beta >0$ must once again be determined so that the first two moments of the approximation match with the true ones.
Prove that the inverse of the distribution function of the random variable is given by
$$
F^{-1}(u) = 1_{{p<u<=1}} :beta^{-1} log(frac{1-p}{1-u})
$$
dirac-delta inverse-function exponential-distribution
asked Dec 22 '18 at 18:04
stedmoaoastedmoaoa
5310
$endgroup$
add a comment |
$begingroup$
We can approximate the conditional distribution by a random variable with density
$$
\pdelta_0(x)+(1-p)beta e^{-beta{x}}1_{x>0}dx
$$
that is to say a weighted mean between a Dirac mass at 0 and an exponential random variable with parameter $beta$. The constants $p in (0,1)$ and $beta >0$ must once again be determined so that the first two moments of the approximation match with the true ones.
Prove that the inverse of the distribution function of the random variable is given by
$$
F^{-1}(u) = 1_{{p<u<=1}} :beta^{-1} log(frac{1-p}{1-u})
$$
dirac-delta inverse-function exponential-distribution
asked Dec 22 '18 at 18:04
stedmoaoastedmoaoa
5310
$endgroup$
add a comment |
$begingroup$
We can approximate the conditional distribution by a random variable with density
$$
\pdelta_0(x)+(1-p)beta e^{-beta{x}}1_{x>0}dx
$$
that is to say a weighted mean between a Dirac mass at 0 and an exponential random variable with parameter $beta$. The constants $p in (0,1)$ and $beta >0$ must once again be determined so that the first two moments of the approximation match with the true ones.
Prove that the inverse of the distribution function of the random variable is given by
$$
F^{-1}(u) = 1_{{p<u<=1}} :beta^{-1} log(frac{1-p}{1-u})
$$
dirac-delta inverse-function exponential-distribution
asked Dec 22 '18 at 18:04
stedmoaoastedmoaoa
5310
$endgroup$
We can approximate the conditional distribution by a random variable with density
$$
\pdelta_0(x)+(1-p)beta e^{-beta{x}}1_{x>0}dx
$$
that is to say a weighted mean between a Dirac mass at 0 and an exponential random variable with parameter $beta$. The constants $p in (0,1)$ and $beta >0$ must once again be determined so that the first two moments of the approximation match with the true ones.
Prove that the inverse of the distribution function of the random variable is given by
$$
F^{-1}(u) = 1_{{p<u<=1}} :beta^{-1} log(frac{1-p}{1-u})
$$
dirac-delta inverse-function exponential-distribution
dirac-delta inverse-function exponential-distribution
asked Dec 22 '18 at 18:04
stedmoaoastedmoaoa
5310
asked Dec 22 '18 at 18:04
stedmoaoastedmoaoa
5310
asked Dec 22 '18 at 18:04
stedmoaoastedmoaoa
5310
asked Dec 22 '18 at 18:04
stedmoaoastedmoaoa
5310
asked Dec 22 '18 at 18:04
stedmoaoastedmoaoa
5310
5310
add a comment |
add a comment |
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