Can X_n Converge to X in Probability if it Doesn't Meet Definitions 2 or 3?

[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).

 

[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) ?

 

[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)

 

[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.

 

[Zone Ranger]

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

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

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

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.

 

[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.

 

[kleinwolf]

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

 

[Zone Ranger]

a.s.=Almost surely

http://www.answers.com/topic/convergence-of-random-variables

 

[Zaare]

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

 

[Zaare]

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

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.

 

[Zone Ranger]

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.

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

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