conditional distribuition and density

[claudiocamera]

How to calculate these conditionals below :
F(X | X>x ) ; F(X | X>x + a ) ; f(X | X>x )

Are all of them zero ?

 

[LouisSheffield]

Before diving into this one, please define (in the context of YOUR text) the precise definition of F(X).

(in particular, how the equality part plays into P(X))

 

[claudiocamera]

X with capital letter is a Random Variable.

F(x) is the probability distribution function and f(x) is the probability density function of the Random Variable X.

F (X|X>x) is the conditional probability distribution function of the Random Variable X, given that X>x.

f(X|X>x) is the conditional density probability function of the Random Variable X given that X>x.

Of course F(x) = P( X< x). Where P means probability.

Actually, there was a mistake in the capital letters, Im gonna correct it now:

F(x | X>x ) ; F(x | X>x + a ) ; f(x | X>x )

All definitions given, I would really appreciate your answer.

Thanks

 

[LouisSheffield]

Thanks - the middle one was the key:
Some texts use F(x)=P(X<x), whereas others use F(x)=P(X≤x).
And thanks for the case adjustment - it had me confused.

The first one is clearly zero, because F(x | X>x) means P(X<x | X>x) / P(X>x), which is 0 / whatever.

The second one depends on the value for a.
If a≥0, we have P(X<x | X>x+a) / P(X>x+a), and we're back to the problem above, except the denominator is a different "whatever" (the numerator's still 0).

However, for a<0 there is some overlap among {X<x, X>x+a}.
That result is P(X<x)/[1 - P(X≤x+a)] = F(x)/[1 - F(x+a)] (I think).

I "think" the third one is f(x) / [1 - F(x)], but it has been sooooo darn long ...

Anyway, hope this helps, especially with considering the value of a.

 

[claudiocamera]

However, for a<0 there is some overlap among {X<x, X>x+a}.
That result is P(X<x)/[1 - P(X≤x+a)] = F(x)/[1 - F(x+a)] (I think)..

I think here the result here is : P(x-|a|<X<x)/[1 - P(X≤x-|a|)] = F(x)- F(x-|a|)/[1 - F(x-|a|)]

I "think" the third one is f(x) / [1 - F(x)], but it has been sooooo darn long ...

Anyway, hope this helps, especially with considering the value of a.

I believe this one is also 0, since F(x |X>x) = 0.

Am I correct ?

 

[LouisSheffield]

Should the numerator be [F(x)-F(x-|a|)]?

 

[claudiocamera]

Should the numerator be [F(x)-F(x-|a|)]?

I guess so, since, when a is negative, the interval under consideration is [ x-|a| ; x] , so values of x less than x-|a| is out of this interval. But I would like to have this confirmed, what do you think ?

 

[LouisSheffield]

The use of |a| certainly changes the problem
(i.e. there's only a single answer now),
but I'll have to get back to you later - work duties call.

Separately, I presume you're doing this for a University class, as opposed to just for fun.

My notation has been sloppy with respect to limits.
There are many cases where F(x) can have impulses, such as the rolling a die.
In such cases, there can be a vast distinction between, say, F(a) and F(a-).
This is much like
P(X>x)!= 1-P(X<x), but rather P(X>x)=1-P(X<=x).
Please check with your professor as to your requirements for formality in these, and similar, limit cases.

Added after 4 hours 56 minutes:

When you introduced |a|, you removed the ambiguity as to there being two solutions.

If the restated problem were F[ x | X>(x+|a|) ], then the result would be zero, much like problem 1.

However, you've expressed it just above as
F[ x | X>(x-|a|) ], which does have a region of overlap, depending on |a|.

I think the problem becomes
F[ x| X>(x-|a|) ] = P[ X<x | X>(x-|a|) ] / P[ X>(x-|a|) ] = P[ X<x | X>(x-|a|) ] / (1 - P[ X<=(x-|a|) ]
so, yes ... the answer should be [ F(x) - F(x-|a|) ] / [ 1 - F(x-|a|) ]

As always, all help robustly correct, except where fatally flawed.

 

[claudiocamera]

When I introduced |a|, I considered only a<0 so that x+a became x-|a| just for this condition.

I think that in spite of it is not a usual mathematical notation, it is correct. It was the better way I came across to understandind and showing the situation.

 

[LouisSheffield]

I meant it as nothing unusual, and it is better in that it isolates the single aspect of the problem.
Nothing was meant whatsoever in any derogatory fashion

Added after 1 minutes:

The "fatally flawed" reference was to me, and having not done much with this stuff for the past 25 years

 

[claudiocamera]

I just explain what I have done, I wrote that it was unusual, since I have never seen this kind of approach in books. I have never thought you were intend to do any negative comment. Actually I don't know what is the meaning of the expression "fatally flawed" I will look up in the dictionary.... Besides statistics, you are also helpping me in my English skills.

Thanx