Question

2. Let Xn ~ NG, Intuitivel y, Xn will concentrate at as n -o. In this question, we will justify this intuition using the convergence concepts we learned (a) Show that Xn, 4> 1/2. (Recall for a random variable X which takes value 1/2 with all probability, its c.d.f. Fo is given by Fo(t) 0 for all t< 1/2, Fo(t)1 for all t 2 1/2. You need then to show the c.d.f. of Xn, say F(t), converges to Fo(t) at all continuous points of Fo.) (b) Show that X 1/2.

Answer 1

This is a simple problem related to convergence of radom variables and their probabilities.

We shall take the random variable distribution

Xn~N(1/2,1/n)

This distribution has a mean /expected value of 1/2 and variance of 1/n (dependent on n =sample size)

Now, our strategy to solve this problem will be to check for the convergence and variations in close proximity of the mean (=1/2) and we will observe its nature and behaviour and then we shall generalize it.

Let F be the distribution function for a point mass at 1/2.

Let us take a modified distribution √ nXn .

Clearly its standard deviation will be 1 (√ n.1/√ n).

Let Z denote a standard normal random variable representing the new distirbution √ nXn

Thus for t < 1/2,

Then as per our standard definition of F distribution

Clearly if we multiply each variable with √ n and hence for all t ,the same inequality holds for the f distribution

Hence

or, $F_{n}(t) =P(\sqrt(n)X_{n}< \sqrt(n)t)$

or, $F_{n}(t) =P( Z< \sqrt(n)t)$

Now think !

what happen of this when n tends to infinity?

Clearly √ n .t will tend to infinity since n will tend to be a very large number and hence we can see that Z will also tend to inifinity and the probability of it occuring will hence tend towards 0.

And hence

$F_{n}(t) =P( Z< \sqrt(n)t)\rightarrow 0$

Now what does this mean ?

This means that that a small change to the left of 1/2 on the numberline will have a ZERO effect on the overall change.

Now let us check for the right hand tendency at 1/2

we will do the same process

For t>1/2

or, $F_{n}(t) =P(\sqrt(n)X_{n}< \sqrt(n)t)$

or, $F_{n}(t) =P( Z< \sqrt(n)t)$

since $n \rightarrow \infty$ , $\sqrt{n} t \rightarrow \infty$

$F_{n}(t) =P( Z< \sqrt(n)t)\rightarrow 1$

Hence from the left and right variable variations we can now say that

$F_{n}(t) \rightarrow F(t)$ for $t \neq 1/2$

and thinking conversely in terms of Xn which defines the Fn distribution we can now say that if

$F_{n}(t) \rightarrow F(t)$  

then $X_{n} \rightarrow Mean value$

or, $X_{n} \rightarrow 1/2$

or $X_{n} \overset{d}{\rightarrow} 1/2$

Convergence of the probability

We will take the help of Markov's inequality to prove this.

For any $\epsilon >0$

Let us check

$P(|X_{n}-X|>\epsilon )$

$P(|X_{n}-X|>\epsilon )=P(|X_{n}-X|^2>\epsilon^2 )$

But by Markov's inequality

$P(|X_{n}-X|^2>\epsilon^2 ) \leq E(|X_{n}-X|^2) /\epsilon^2$

thus we have

$P(|X_{n}-X|>\epsilon )=P(|X_{n}-X|^2>\epsilon^2 )$

$P(|X_{n}-X|>\epsilon ) \leq E(|X_{n}-X|^2) /\epsilon^2$

Now as $n \rightarrow \infty$

$P(|X_{n}-X|>\epsilon ) \leq E(|X_{n}-X|^2) /\epsilon^2 \rightarrow 0$

So what do we infer ?

We infer that the probability of any variation /divergence from the given variable is 0

and hence we can say that the probability of the distribution tends to converge to the distribution of probablity of the sample distirbution.

Xn converges to X in probability

or, $X_{n} \overset{P}{\rightarrow} X$

but X=1/2 for the given sample

thus $X_{n} \overset{P}{\rightarrow} 1/2$