Family of densities on [0,1] with strictly positive density at zero
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I am looking for a parameterized family of probability densities on the unit interval, which all share a strictly positive density at zero. Ideally, they could be parameterized by some parameter $lambda$, such that a higher $lambda$ implies first order stochastic dominance. Is there something like this, or maybe close to it?
Thanks!
probability integration probability-distributions
edited Jan 18 at 10:14
md2perpe
8,80411129
asked Jan 15 at 0:55
user509037user509037
1006
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add a comment |
$begingroup$
I am looking for a parameterized family of probability densities on the unit interval, which all share a strictly positive density at zero. Ideally, they could be parameterized by some parameter $lambda$, such that a higher $lambda$ implies first order stochastic dominance. Is there something like this, or maybe close to it?
Thanks!
probability integration probability-distributions
edited Jan 18 at 10:14
md2perpe
8,80411129
asked Jan 15 at 0:55
user509037user509037
1006
$endgroup$
add a comment |
$begingroup$
I am looking for a parameterized family of probability densities on the unit interval, which all share a strictly positive density at zero. Ideally, they could be parameterized by some parameter $lambda$, such that a higher $lambda$ implies first order stochastic dominance. Is there something like this, or maybe close to it?
Thanks!
probability integration probability-distributions
edited Jan 18 at 10:14
md2perpe
8,80411129
asked Jan 15 at 0:55
user509037user509037
1006
$endgroup$
I am looking for a parameterized family of probability densities on the unit interval, which all share a strictly positive density at zero. Ideally, they could be parameterized by some parameter $lambda$, such that a higher $lambda$ implies first order stochastic dominance. Is there something like this, or maybe close to it?
Thanks!
probability integration probability-distributions
probability integration probability-distributions
edited Jan 18 at 10:14
md2perpe
8,80411129
asked Jan 15 at 0:55
user509037user509037
1006
edited Jan 18 at 10:14
md2perpe
8,80411129
asked Jan 15 at 0:55
user509037user509037
1006
edited Jan 18 at 10:14
md2perpe
8,80411129
edited Jan 18 at 10:14
md2perpe
8,80411129
edited Jan 18 at 10:14
md2perpe
8,80411129
8,80411129
asked Jan 15 at 0:55
user509037user509037
1006
asked Jan 15 at 0:55
user509037user509037
1006
asked Jan 15 at 0:55
user509037user509037
1006
1006
add a comment |
add a comment |
1 Answer
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You could form a density like this using a mixture of Beta distributions. For example, we could take a mixture of a uniform distribution (to give non-zero density at all points) and another Beta distribution parameterised with $lambda$ as the mean. This gives the general form:
$$f_X(x) = phi + (1-phi) cdot frac{x^{lambda kappa-1} (1-x)^{(1-lambda) kappa -1}}{B(lambda kappa, (1-lambda) kappa)} quad quad quad text{for all } 0 leqslant x leqslant 1.$$
where $0 < phi < 1$ and $kappa >0$. This distribution has non-zero density at all points and higher values of $lambda$ give higher values (in the sense of first-order stochastic dominance). In particular, it has mean:
$$mathbb{E}(X) = frac{phi}{2} + (1-phi) lambda.$$
answered Jan 15 at 3:31
BenBen
1,930215
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
You could form a density like this using a mixture of Beta distributions. For example, we could take a mixture of a uniform distribution (to give non-zero density at all points) and another Beta distribution parameterised with $lambda$ as the mean. This gives the general form:
$$f_X(x) = phi + (1-phi) cdot frac{x^{lambda kappa-1} (1-x)^{(1-lambda) kappa -1}}{B(lambda kappa, (1-lambda) kappa)} quad quad quad text{for all } 0 leqslant x leqslant 1.$$
where $0 < phi < 1$ and $kappa >0$. This distribution has non-zero density at all points and higher values of $lambda$ give higher values (in the sense of first-order stochastic dominance). In particular, it has mean:
$$mathbb{E}(X) = frac{phi}{2} + (1-phi) lambda.$$
answered Jan 15 at 3:31
BenBen
1,930215
$endgroup$
add a comment |
$begingroup$
You could form a density like this using a mixture of Beta distributions. For example, we could take a mixture of a uniform distribution (to give non-zero density at all points) and another Beta distribution parameterised with $lambda$ as the mean. This gives the general form:
$$f_X(x) = phi + (1-phi) cdot frac{x^{lambda kappa-1} (1-x)^{(1-lambda) kappa -1}}{B(lambda kappa, (1-lambda) kappa)} quad quad quad text{for all } 0 leqslant x leqslant 1.$$
where $0 < phi < 1$ and $kappa >0$. This distribution has non-zero density at all points and higher values of $lambda$ give higher values (in the sense of first-order stochastic dominance). In particular, it has mean:
$$mathbb{E}(X) = frac{phi}{2} + (1-phi) lambda.$$
answered Jan 15 at 3:31
BenBen
1,930215
$endgroup$
add a comment |
$begingroup$
You could form a density like this using a mixture of Beta distributions. For example, we could take a mixture of a uniform distribution (to give non-zero density at all points) and another Beta distribution parameterised with $lambda$ as the mean. This gives the general form:
$$f_X(x) = phi + (1-phi) cdot frac{x^{lambda kappa-1} (1-x)^{(1-lambda) kappa -1}}{B(lambda kappa, (1-lambda) kappa)} quad quad quad text{for all } 0 leqslant x leqslant 1.$$
where $0 < phi < 1$ and $kappa >0$. This distribution has non-zero density at all points and higher values of $lambda$ give higher values (in the sense of first-order stochastic dominance). In particular, it has mean:
$$mathbb{E}(X) = frac{phi}{2} + (1-phi) lambda.$$
answered Jan 15 at 3:31
BenBen
1,930215
$endgroup$
You could form a density like this using a mixture of Beta distributions. For example, we could take a mixture of a uniform distribution (to give non-zero density at all points) and another Beta distribution parameterised with $lambda$ as the mean. This gives the general form:
$$f_X(x) = phi + (1-phi) cdot frac{x^{lambda kappa-1} (1-x)^{(1-lambda) kappa -1}}{B(lambda kappa, (1-lambda) kappa)} quad quad quad text{for all } 0 leqslant x leqslant 1.$$
where $0 < phi < 1$ and $kappa >0$. This distribution has non-zero density at all points and higher values of $lambda$ give higher values (in the sense of first-order stochastic dominance). In particular, it has mean:
$$mathbb{E}(X) = frac{phi}{2} + (1-phi) lambda.$$
answered Jan 15 at 3:31
BenBen
1,930215
answered Jan 15 at 3:31
BenBen
1,930215
answered Jan 15 at 3:31
BenBen
1,930215
answered Jan 15 at 3:31
BenBen
1,930215
1,930215
add a comment |
add a comment |
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