Determine whether the PDF converges in distribution

In summary: 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:$$ \lim_{n\to\infty} F_{X_n}(x) = F(x) \quad\forall x\in\mathbb{R} $$$$ \iff \lim_{n\to\infty} \mathbb{P}\big( X_n \leq x \big) = \mathbb{P}\big( X \leq x \big) \quad\forall x\in\mathbb{R} $$
  • #1
cwbullivant
60
0

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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  • #2
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).
 
  • #3
Edit: Ok, I think I've written too much. First follow the hint of the post above.
 
  • #4
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.
 
  • #5
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!
 
  • #6
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).
 
  • #7
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}##.
 
  • #8
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:
  • #9
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##?
 

1. What does it mean for a PDF to converge in distribution?

Convergence in distribution refers to the behavior of a sequence of probability density functions (PDFs) as the number of terms in the sequence approaches infinity. If the sequence converges, it means that the shape of the PDFs becomes increasingly similar as more terms are added.

2. How is convergence in distribution different from convergence in probability?

Convergence in distribution is a weaker form of convergence compared to convergence in probability. While convergence in probability requires that the probability of the difference between the sequence of random variables and the limit approaches zero, convergence in distribution only requires that the cumulative distribution functions (CDFs) of the sequence of random variables converge to the CDF of the limit.

3. What is the relationship between convergence in distribution and the central limit theorem?

The central limit theorem states that the sum of a large number of independent and identically distributed random variables will be approximately normally distributed. This is related to convergence in distribution because as the number of terms in the sequence of random variables increases, the shape of the PDF will tend towards a normal distribution, demonstrating convergence in distribution.

4. How can I determine if a PDF converges in distribution?

To determine if a PDF converges in distribution, you can use the definition of convergence in distribution and compare the CDF of the sequence of random variables to the CDF of the limit. If they become increasingly similar as the number of terms in the sequence increases, then the PDF converges in distribution.

5. What are some real-world applications of studying convergence in distribution?

Convergence in distribution has many practical applications in fields such as statistics, economics, and finance. It is used to understand the behavior of random variables and the limiting behavior of certain models. For example, it is used in studying stock market trends and forecasting future values.

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