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Re: Re: indeterminate and infinity expressions

• To: mathgroup at smc.vnet.net
• Subject: [mg67699] Re: [mg67615] Re: indeterminate and infinity expressions
• From: Søren Merser <merser at image.dk>
• Date: Wed, 5 Jul 2006 04:17:30 -0400 (EDT)
• References: <e85es4$k8e$1@smc.vnet.net> <200607021027.GAA08948@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Hi
First my most sincerly apologies for the typos!
I was unsuccefully trying to make my problem clearer.
I'll try to rephrase it:
Want to calculate the residaul deviance according to the following formula:

Some data:
dep = {1, 0, 1, 0}; (* vector with fitted values *)
yhat = {.012, .123, .23, .563}; (* vector with fitted values *)

Problem:

Sign[dep-yhat] Sqrt[dep Log[dep/yhat] + (dep-yhat) Log[(1-dep)/(1-yhat)]]

My solution:

expr1 = dep Log[dep/yhat] /. Indeterminate -> 0;
expr2 = (1 - dep) Log[(1 - dep)/(1 - yhat)] /. Indeterminate -> 0;
Sign[dep - yhat] Sqrt[2(expr1 + expr2)]


Regards Soren

----- Original Message ----- 
From: "David W.Cantrell" <DWCantrell at sigmaxi.org>
To: mathgroup at smc.vnet.net
Subject: [mg67699] [mg67615] Re: indeterminate and infinity expressions


> Søren_Merser <merser at image.dk> wrote:
>> I need to evaluate the expression below:
>>
>> In[1]:= Log[(#/yhat)] + (1-#) Log[(1-#)/(1-yhat)] &  /@  {0,1} /.
>> yhat->123 Out[1]= {Indeterminate, -Infinity}
>
> First, Søren, if you really want others to help you, please be much more
> careful in your posting. There were several errors.
>
> Evaluating your In[1], exactly as you gave it, returns
>
> {-Infinity, Indeterminate}, which is perfectly correct.
>
> Your claimed Out[1] had the list in the wrong order.
>
> But looking at your In[2], which is missing /. BTW, it seems that your
> In[1] was missing an initial # and that yhat was to have been replaced
> by .123, rather than 123 .
>
> To get what you want, use Limit. Here's my suggestion:
>
> In[8]:= Limit[x Log[(x/yhat)] + (1-x) Log[(1-x)/(1-yhat)], x->#]
> & /@ {0,1} /. yhat -> .123
>
> Out[8]= {0.131248,2.09557}
>
> David
>
>> Depending on the value of #, I get Infinity and Inderterminate results.
>>
>> To get the wanted answer I've to split the expression into two parts
>>
>> In[2]:= ans1 = # Log[#/yhat]  &  /@ {0, 1} yhat -> .123
>> Out[2]= {Indeterminate, 2.09557}
>>
>> In[3]:= ans2 = (1-#) Log[(1-#)/(1-yhat)] &  /@ {0, 1} /. yhat  -> .123
>> Out[3]= {0.131248, Indeterminate}
>>
>> In[4]:= Max[#] & /@ ({ans1, ans2} /. Indeterminate -> 0)
>> Out[4]= {2.09557, 0.131248}
>>
>> Now, is there an easier way to do it?
>>
>> Regards Soren
> 

• References:
  • Re: indeterminate and infinity expressions
    • From: "David W.Cantrell" <DWCantrell@sigmaxi.org>