Limit definition examples

1. Mar 10, 2005

Zaare

First 3 definitions:
$$\begin{array}{l} \left( 1 \right)\mathop {\lim }\limits_{n \to \infty } P\left( {\left| {X_n - X} \right| > \varepsilon } \right) = 0 \\ \left( 2 \right)P\left( {\mathop {\lim }\limits_{n \to \infty } X_n = X} \right) = 1 \\ \left( 3 \right)\mathop {\lim }\limits_{n \to \infty } E\left[ {\left( {X_n - X} \right)^2 } \right] = 0 \\ \end{array}$$

I need to find:
a. an example that (1) does not give (3).
b. an example that (1) does not give (2).

2. Mar 28, 2005

kleinwolf

Your stuff seems quite hard...I don't know where to begin...are you sure there are no things like 2=>1, or other logical implications ??

Have you tried some series like : X_n=1/ln(n) ?

3. Mar 28, 2005

Zone Ranger

let X_1 =1 on [0,1]
let X_2=1 on [0,1/2] , 0 otherwise
let x_3=1 on [1/2,1] , 0 ow
let x_4=1 on[0,1/3] ,0 ow
let x_5=1 on [1/3,2/3] , 0 ow
....

can you show that X_n converges in probability (1)
but x_n does not converge a.s. (2)

4. Mar 28, 2005

kleinwolf

Hm..Zone Ranger interpreted the x_n as functions....I took them as numbers.

If X_n are functions, then it's easy to find what you want:

Let X_n be functions over [0;1], with

X_n(x)=g(n) if x in [0;1/n], 0 otherwise, with a stricly increasing function g(n)

Then

1) P(|X_n|>e)<=1/n, hence the limit gives 0
b: 2) X_n does not tend towards the 0 function, since X_n(0)>0 forall n
a: E(X_n^2)=g(n)^2/n...here choose g(n)=Sqrt(n)....you get

limit n->infty E((X_n-0)^2)=1

But if the X_n are numbers, I don't know how to solve this...

Thanks to Zone Ranger.

5. Mar 28, 2005

Zone Ranger

the X_n have to be random variables (measurable functions).

you are correct that for your choice of X_n, X_n(0)>0. but still X_n->0 a.s.
so with your X_n (2) still holds.

6. Mar 28, 2005

BicycleTree

Assuming the X_n's do not have be a random sample from X, then you could define the RV's as discrete, each taking on a value with probability 1 and all other values with probability 0. It's easy to set up examples that way.

Last edited: Mar 28, 2005
7. Mar 28, 2005

kleinwolf

I don't know what u mean with a.s. (maybe converges uniformly ???)

8. Mar 28, 2005

9. Apr 7, 2005

Zaare

I'm sorry for not answering sooner, but after 10 days and no answer I thought noone was interested.

I know that (2) gives (1) and also (3) gives (1), these two are not hard to prove.
I'm going to try your suggestions now.

10. Apr 7, 2005

Zaare

Taking this example, if I understand correctly, X_n does not converge almost surely (2) since it does not converge to a single value but rather keeps dividing the interval [0,1] into halfs, and "jumps back and forth" within the interval.
X_n does not converge in mean square to X (3) because of the same reason, namely the expected value of $$(X_n-X)^2$$ kepps changing as $${n \to \infty }$$.
However, I don't know how to show that X_n converges in probability.

11. May 2, 2005

Zone Ranger

$$X_n$$ does converge in mean square to X (X=0)

the expected value of $$(X_n-X)^2$$ goes to 0 as $${n \to \infty }$$.