Determine whether the PDF converges in distribution

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Homework Statement



Let $$ \it{f}(x) $$ be a probability density function. Now let Xn have the density:

$$ \it{f}_{n}(x) = n\it{f}(nx) $$

Determine whether or not Xn converges in distribution to zero.

(this is the verbatim statement, there is no additional information given)

Homework Equations


[/B]
If the CDF $$ F_{n}(X) \rightarrow 0, n \rightarrow \infty $$ then Xn converges to zero in distribution.

The Attempt at a Solution



Definition of CDF:

$$ F_{n}(x) = P (X < x) = \int_{-\infty}^{x} \it{f}_{n}(x) dx= \int_{-\infty}^{x} n\it{f}(nx) dx $$

And to be a valid PDF, its integral from -∞ to +∞ must be 1. That requires that f(nx) must go to zero as nx goes to ±∞. So the lower limit on that integral is already taken care of; that part has to vanish if nf(nx) is going to be a valid PDF.

But for finite x, wouldn't the behavior of f(nx) as n goes to ∞ depend on exactly what f(nx) looks like (e.g. would go to zero if f(nx) = e^{-nx}, but not if f(nx) = ne^(-x))? I'm not sure on where to go from here.
 
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cwbullivant said:

Homework Statement



Let $$ \it{f}(x) $$ be a probability density function. Now let Xn have the density:

$$ \it{f}_{n}(x) = n\it{f}(nx) $$

Determine whether or not Xn converges in distribution to zero.

(this is the verbatim statement, there is no additional information given)

Homework Equations


[/B]
If the CDF $$ F_{n}(X) \rightarrow 0, n \rightarrow \infty $$ then Xn converges to zero in distribution.

The Attempt at a Solution



Definition of CDF:

$$ F_{n}(x) = P (X < x) = \int_{-\infty}^{x} \it{f}_{n}(x) dx= \int_{-\infty}^{x} n\it{f}(nx) dx $$

And to be a valid PDF, its integral from -∞ to +∞ must be 1. That requires that f(nx) must go to zero as nx goes to ±∞. So the lower limit on that integral is already taken care of; that part has to vanish if nf(nx) is going to be a valid PDF.

But for finite x, wouldn't the behavior of f(nx) as n goes to ∞ depend on exactly what f(nx) looks like (e.g. would go to zero if f(nx) = e^{-nx}, but not if f(nx) = ne^(-x))? I'm not sure on where to go from here.

Re-write ##F_n(x)## by changing variables to ##u = nt## in ##\int_{-\infty}^{x} n f(nt)\, dt##. Note: do NOT ##\int^x f(x) dx##; you should never, ever use the same symbol '##x##' to stand for two totally different things in the same expression---once as the integration variable, and once as the upper integration limit).
 
Edit: Ok, I think I've written too much. First follow the hint of the post above.
 
cwbullivant said:
Definition of CDF:

$$ F_{n}(x) = P (X < x) = ... $$

This is wrong. A CDF is actually defined as ##F(x) = P\big( X\leq x \big)##.

It's a technical but important distinction. A CDF is alway right continuous . This matters when you start dealing with CDFs involving discrete r.v.'s, hybrids, etc.
 
StoneTemplePython said:
This is wrong. A CDF is actually defined as ##F(x) = P\big( X\leq x \big)##.

It's a technical but important distinction. A CDF is alway right continuous . This matters when you start dealing with CDFs involving discrete r.v.'s, hybrids, etc.

Some books and papers (mostly old, now) actually define the CDF using "<". Probably 99.9% of sources nowadays use "≤", but I am not sure we can positively assert that "<" is wrong. (Maybe the OP's textbook and/or course notes take the "<" definition---unlikely, I know, but just barely possible.) However, I realize that many students are sloppy, and write "<" when they really mean "≤", so pointing it out to them is good!
 
Ray Vickson said:
I realize that many students are sloppy, and write "<" when they really mean "≤", so pointing it out to them is good!

Sloppiness is exactly the issue here.

Anyhow, I make the appropriate substitution, with u = nt, du = n dt, and come out to:

$$ \int_{-infty}^{nx} \it{f}(u) du $$

This has eliminated the linear dependence on n from the original statement and reduced the problem to an integral with just a function in the integrand. From here, I don't seem to see much improvement. Whether or not the outcome of the integral goes to zero as n goes to infinity still seems like it should depend on the actual structure of f(u).

It was tempting to say that since the upper bound is u=nx, which is linear in n (which generally doesn't bode well for a finite answer as n goes to infinity), the PDF does not converge in distribution, but that doesn't sound right to me, because I don't see a good way to make that kind of conclusion from the integral of a completely general f(u).
 
cwbullivant said:
Whether or not the outcome of the integral goes to zero as n goes to infinity still seems like it should depend on the actual structure of f(u).

You can deduce the behavior of ##f(y)## for ##y \rightarrow \infty##, namely ##f## should decrease faster than ##\frac{1}{y}##.
 
cwbullivant said:
Sloppiness is exactly the issue here.

Anyhow, I make the appropriate substitution, with u = nt, du = n dt, and come out to:

$$ \int_{-infty}^{nx} \it{f}(u) du $$

This has eliminated the linear dependence on n from the original statement and reduced the problem to an integral with just a function in the integrand. From here, I don't seem to see much improvement. Whether or not the outcome of the integral goes to zero as n goes to infinity still seems like it should depend on the actual structure of f(u).

It was tempting to say that since the upper bound is u=nx, which is linear in n (which generally doesn't bode well for a finite answer as n goes to infinity), the PDF does not converge in distribution, but that doesn't sound right to me, because I don't see a good way to make that kind of conclusion from the integral of a completely general f(u).

You are having trouble because you are using the wrong convergence criterion! Your stated criterion would be for ##X_n \to +\infty##, not the ##X_n \to 0## posed in the question.

Recall the true definition: ##X_n \to X## in distribution if ##F_n(x) \to F(x)## for all ##x## (or, at least, for most ##x##). What is the CDF ##F(x)## for ##X = 0## (the identically-zero random variable)?
 
Last edited:
Ray Vickson said:
What is the CDF ##F(x)## for ##X = 0## (the identically-zero random variable)?

## F(x) = P(0 \leq x) = 1, x \geq 0, 0 otherwise##?
 
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