probability = 1/infinite ?

[Roy]

Let us define a function f (x, y) = xy where x and y can be any nonzero
numbers in the real domain R, e.g., f(2, -3.5) = -7. In other words,
if x is fixed, and y variates, f(x,y) has infinite number of possible
values.

Now assume the value of f(x, y) is fixed, but y can change, can we
prove that any nonzero x in R is equally likely to be mapped to f(x,y)?
How?

My feeling is that the "probability" that x is mapped to a given f(x,y)
is 1/infinite because
1) any x can be mapped to infinite number of f(x,y) when y is randomly
chosen from R
2) for any x there is only one y that can give the fixed value f(x,y)

However, it seems weird if the probability = 1/infinite = 0. Looks
like I am talking about the probability of an individual outcome of a
continuous random variable. I got confused.

Any comments/thoughts?

Thanks a lot
Roy

[Robert Israel]

In article <1137644093....@g49g2000cwa.googlegroups.com>,

Roy <roy...@hotmail.com> wrote:
>Let us define a function f (x, y) = xy where x and y can be any nonzero
>numbers in the real domain R, e.g., f(2, -3.5) = -7. In other words,
>if x is fixed, and y variates, f(x,y) has infinite number of possible
>values.
>
>Now assume the value of f(x, y) is fixed, but y can change, can we
>prove that any nonzero x in R is equally likely to be mapped to f(x,y)?

This statement is meaningless as it stands.

>How?
>
>My feeling is that the "probability" that x is mapped to a given f(x,y)
>is 1/infinite because
>1) any x can be mapped to infinite number of f(x,y) when y is randomly
>chosen from R
>2) for any x there is only one y that can give the fixed value f(x,y)
>
>However, it seems weird if the probability = 1/infinite = 0. Looks
>like I am talking about the probability of an individual outcome of a
>continuous random variable. I got confused.

Yes, you did. First of all, "randomly chosen from R" has no meaning
unless you specify a probability distribution. If you fix a nonzero
value k for the product and you choose Y from a continuous distribution,
then X = k/Y also has a continuous distribution, and so the probability
that X takes any particular value is 0.

Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[Roy]

Thank you very much.
However, intuitively, it seems each outcome of x is equally likely to
be mapped to k for the product.
Is there any way to prove such claim? Assume random variable y has a
uniform distribution in R.

[Robert Israel]

There is no such thing as a uniform distribution on the whole real
line.
You could take the next-best thing as a sequence of distributions that
approach uniformity in the sense that the ratios of densities
f(a)/f(b) -> 1 for every a and b. But if you take such a sequence of
distributions for Y, the corresponding distributions for X will not
approach uniformity.

If Y has a distribution with density f_Y, then X = k/Y has density
f_X(x) = k/x^2 f_Y(k/x). So if f_Y(a)/f_Y(b) -> 1 for every a and b,
f_X(a)/f_X(b) -> b^2/a^2.

[David C. Ullrich]

There's no such thing as a uniform distribution on R. (That's why
he said you needed to specify a disytibution - if there were such
a thing as a uniform distribution on R then people would assume
that that was what was meant when someone said something about
a "random" real number.)

************************

David C. Ullrich

[Roy]

Yes, I totally agree with you.
But even if f_X(a)/f_X(b) -\-> 1, can we still say something about
f(X=a --> XY=k) / f(X=b --> XY=k)? where f(X=a ---> XY=k) is a kind of
transition probability or density.
Can we say this kind of transition (or mapping) has a "uniform
distribution" or is "equally likely".

Thanks
Roy

[Roy]

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