Re: Bug with ProbabilityDistribution
- To: mathgroup at smc.vnet.net
- Subject: [mg132542] Re: Bug with ProbabilityDistribution
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Thu, 10 Apr 2014 03:07:37 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-outx@smc.vnet.net
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- References: <20140409081446.CA0646A01@smc.vnet.net>
Works as expected if you simplify the Piecewise (PiecewiseExpand) or let
the Piecewise default handle the cases where the value is zero. Note that
PDF[d][{x1,x2}] can be entered simply as PDF[d,{x1,x2}]
d = ProbabilityDistribution[
Piecewise[{
{0, x1 == 1 && x2 == 1},
{1/2, x1 == 1 && x2 == 0},
{0, x1 == 0 && x2 == 1},
{1/2, x1 == 0 && x2 == 0}}] //
PiecewiseExpand,
{x1, 0, 1, 1}, {x2, 0, 1, 1}];
t1 = Table[
{x1, x2, PDF[d, {x1, x2}], PDF[d][{x1, x2}]},
{x1, 0, 1}, {x2, 0, 1}] //
Flatten[#, 1] &
{{0, 0, 1/2, 1/2}, {0, 1, 0, 0}, {1, 0, 1/2, 1/2},
{1, 1, 0, 0}}
d = ProbabilityDistribution[
Piecewise[{
{1/2, x1 == 1 && x2 == 0},
{1/2, x1 == 0 && x2 == 0}}],
{x1, 0, 1, 1}, {x2, 0, 1, 1}];
t2 = Table[
{x1, x2, PDF[d, {x1, x2}], PDF[d][{x1, x2}]},
{x1, 0, 1}, {x2, 0, 1}] //
Flatten[#, 1] &
{{0, 0, 1/2, 1/2}, {0, 1, 0, 0}, {1, 0, 1/2, 1/2},
{1, 1, 0, 0}}
t1 == t2
True
Bob Hanlon
On Wed, Apr 9, 2014 at 4:14 AM, CHARLES GILLINGHAM <cgillingham1 at me.com>wrote:
> Evaluate this:
>
> ProbabilityDistribution[
> Piecewise[{{0, x1 == 1 && x2 == 1}, {1/2, x1 == 1 && x2 ==
> 0}, {0, x1 == 0 && x2 == 1}, {1/2, x1 == 0 && x2 == 0}}],
> {x1, 0, 1, 1}, {x2, 0, 1, 1}
> ]
> PDF[%][{1, 1}]
>
> And you get:
>
> Undefined
>
> But the probability is obviously 0
>
> Is there something I am misunderstanding about how simple Boolean
> probability distributions should be set up?
>
> Thanks.
>
>
>
>
>
- References:
- Bug with ProbabilityDistribution
- From: CHARLES GILLINGHAM <cgillingham1@me.com>
- Bug with ProbabilityDistribution