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Re: PiecewiseExpand bug?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg68500] Re: PiecewiseExpand bug?
*From*: Mark Fisher <mark at markfisher.net>
*Date*: Tue, 8 Aug 2006 06:28:36 -0400 (EDT)
*References*: <eb45vn$n7p$1@smc.vnet.net> <eb6k48$nlh$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Let me put it slightly differently:
expr1 = Product[PDF[BernoulliDistribution[p], x[i]], {i, 3}];
expr2 = PiecewiseExpand[expr1];
Simplify[expr1 == expr2]
produces
p^2 == p || x[1] != 0 || x[2] != 1 || x[3] != 1
so that if
x[1] == 0 && x[2] == 1 && x[3] == 1 && p^2 != p
then expr1 and expr2 are not the same!
Therefore, if PiecewiseExpand is *not* supposed to change the what the
expression evaluates to, then this would seem to be a bug.
--Mark.
Jean-Marc Gulliet wrote:
> Mark Fisher wrote:
>> This produces what I expect:
>>
>> Product[PDF[BernoulliDistribution[p], x[i]], {i, 3}] /.
>> {x[1] -> 0, x[2] -> 1, x[3] -> 1}
>>
>> But this does not:
>>
>> PiecewiseExpand[Product[PDF[BernoulliDistribution[p], x[i]], {i, 3}]] /.
>> {x[1] -> 0, x[2] -> 1, x[3] -> 1}
>>
>> It appears PiecewiseExpand does something it shouldn't here. Any thoughts?
>
> No, it does not: you are not computing the same expressions because of
> the location of the transformation rules. You can clearly see what's
> going on with the Trace function if needed.
>
> In[1]:=
> << "Statistics`DiscreteDistributions`"
>
> In[2]:=
> Product[PDF[BernoulliDistribution[p], x[i]], {i, 3}] /. {x[1] -> 0, x[2]
> -> 1, x[3] -> 1}
>
> Out[2]=
> (1 - p)*p^2
>
> In[3]:=
> PiecewiseExpand[Product[PDF[BernoulliDistribution[p], x[i]], {i, 3}]] /.
> {x[1] -> 0, x[2] -> 1, x[3] -> 1}
>
> Out[3]=
> 0
>
> In[4]:=
> PiecewiseExpand[Product[PDF[BernoulliDistribution[p], x[i]], {i, 3}] /.
> {x[1] -> 0, x[2] -> 1, x[3] -> 1}]
>
> Out[4]=
> (1 - p)*p^2
>
> In[5]:=
> PiecewiseExpand[(1 - p)*p^2]
>
> Out[5]=
> (1 - p)*p^2
>
> In[6]:=
> Trace[Product[PDF[BernoulliDistribution[p], x[i]], {i, 3}] /. {x[1] ->
> 0, x[2] -> 1, x[3] -> 1}]
>
> In[7]:=
> Trace[PiecewiseExpand[Product[PDF[BernoulliDistribution[p], x[i]], {i,
> 3}]] /. {x[1] -> 0, x[2] -> 1, x[3] -> 1}]
>
> Regards,
> Jean-Marc
>
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