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qwertykeyboards
05-19-2010, 05:44 AM
Hi,

the following question keeps coming up in recent year's exam and i'm unsure how to do it fully:

Q. If the sequence of random variables {Xn}n=1 up to infinity, is such that, for each n element of N,

n with probability 1/n
Xn =
0 with probability [1-(1/n)]

(i) Does Xn converge in probability?
(ii) Does Xn converge in L1?

i have done a similar problem and got it right but don't fully know the answer to this one. for (i) i have: when Xn converges to X in prob. lim n->infinity P (mod of Xn-X < epsilon) = 1, for all epsilon > 0. so plugging in Xn = n would give me 0< epsilon, meaning prob of this would have to be 1 since epsilon is greater than 0. the same for Xn = 0. my problem with this is that it doesnt involve the given probabilities 1/n and 1-1/n so i feel theres more to it.

same thing with (ii), where do the probs come into it? convergence in L1 is lim of E[ mod Xn-X] = 0. plugging in Xn = n and Xn = 0 both give 0 so id say that Xn does converge in both probability and in L1.

am i correct? i'm sure they didnt just put the probabilities in the question just for show so i must be wrong.

any help would be greatly appreciated!!! exam tomorrow.

BGM
05-19-2010, 10:18 AM
Below I just try to mimic the arguments in wiki.

Note:
The sequence of random variable \{X_n\}_{n=1}^{\infty}
converge to 0 in distribution and in probability.

1. First, it is not hard to see that
The distribution function of X_n ,
F_n(x) = \begin{matrix} 0 & x < 0\\
1 - \frac {1} {n} & 0 \leq x < n \\
1 & x \geq n \end{matrix}
So the limiting distribution function,
F(x) = \begin{matrix} 0 & x < 0\\
1 & x \geq 0 \end{matrix}
So X_n \xrightarrow{d} 0

Since {X_n} converge to the constant 0 in distribution,
it also converge to the constant 0 in probability

2. \forall \epsilon > 0, \forall \delta > 0
we set N_\delta = \frac {1} {\delta} + 1
such that Pr\{|X_n - 0| \geq \epsilon\}
= Pr\{X_n \geq \epsilon\}
= \begin{matrix} \frac {1} {n} & 0 < \epsilon \leq n \\
0 & \epsilon > n \end{matrix}
\leq \frac {1} {n} < \delta \: \forall n \geq N_\delta
So X_n \xrightarrow{p} 0

3. E[X_n] = n \frac{1}{n} = 1 \: \forall n
So X_n \xrightarrow{L^1} 1