Re: define the division of X/Y


Proginoskes wrote:

schoenfeld.one@xxxxxxxxx wrote:
Proginoskes wrote:
schoenfeld.one@xxxxxxxxx wrote:
Proginoskes wrote:
Mike wrote:
where X and Y are two random variables...

What is the precise and rigorous condition on Y to avoid the "dividing by
zero" problem?

Hmm. Making sure Y is nonzero would do it.

Making the probability that
Y is zero equal to zero might also do it.

That's an insufficient condition.

The probabiltiy of drawing 5 out of the Naturals is 0.

What distribution are you using?

The probability
of drawing a 0.5 out of the set [0,1] is 0. It does not mean that they
cannot ever be drawn.

But things such as the expected value (and moments) can still be
calculated if the probability of dividing by zero is 0.

And I did NOT mean that "probability of event E = 0" is the same as "E
cannot happen". The more precise statement of my intention was: "Y is
nonzero almost everywhere".

If P(Y == 0) = 0 does not imply that X/Y is well defined for all X in
Reals, Y in Reals,

That's a weird implication.

It sure is, and it's all yours.

then how P(Y==0)=0 help with the OP's question?

Because X and Y are random variables, not arbitrary real numbers. This
suggests that the setting is a probability space, not general
arithmetic.

Just because the probability of drawing (y=0) is 0 does not mean that
(y=0) is never drawn. The OP necessarily requires that (y=0) is never
drawn.

The OP (Mike) didn't say exactly what he wanted to do, so all of this
is speculation anyway. But I suspect he might want to calculate the
probability that X/Y is between 2 or 3, or the expected value of X/Y,
or _some_ statistic involving X/Y.

I don't see how you came up with that.

For instance, X and Y might be uniformly distributed over [-1,1], and
Mike wants to calculate the probability that X/Y is between 2 and 3.
The problem is that X/Y is not defined, so you could end up with an
improper integral when trying to calculate this probability.

One solution might be to change the distribution to one of the form:
p(x) = 1/2, if x is not 0;
p(x) = 0, if x = 0.

Then, if Y has this probability distribution, Y will definitely never
be 0. The only concern is that this is a different distribution.
However, since the probability distribution has been changed on a set
of measure 0, you will get the same answer for problems like "What is
the probability that Y > -0.5?" as you would for the old distribution.
However, you can now talk about X/Y, since Y will never be 0.

That's not what the OP asked. The OP asked what were the 'precise and
rigorous conditions' for avoiding a divide by zero for random x and
random y. You said letting P(y==0)=0 was sufficient. I am saying its
insufficient.

.

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