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I understand convergence in probability concept (theoretically), but I'm still having trouble to do this kind of example:

Suppose $X ∼ Uniform[0, 1]$, and let $Y_n = \frac{n−1}{n} \cdot X$

Prove that $Y_n \rightarrow^{P} X$.

I started off by finding:

$E[X_i]=\frac{1}{2}, \quad Var(X_i)=\frac{1}{12}$

$E[Y_n]=\frac{n-1}{n}\cdot E[X]=\frac{n-1}{2n}$

$Var(Y_n)=\frac{(n-1)^2}{n^2}\cdot \frac{1}{12}=\frac{(n-1)^2}{12n^2}$

So then I figure I should straight away plug into:

$\lim_{n\to\infty}P(|Y_n - X|> \epsilon) = \lim_{n\to\infty}\frac{(n-1)^2}{12n^2\epsilon^2} = \frac{1}{12\epsilon^2} \neq0$

I'm pretty sure I did it wrong, but I don't know how to go about from here.

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2 Answers 2

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$ X $ is uniformly distributed on $[0,1]$, $Y_n = \frac{n-1}{n}X$

We want to use Chebyshew inequality : $\Bbb P(|Y_n-X| > \epsilon) \leq \frac{D^2(Y_n-X)}{\epsilon^2} $

So we need to compute that variance.

Firstly $E[Y_n-X] = E[Y_n] - E[X] = \frac{n-1}{2n} - \frac{n}{2n} = \frac{-1}{2n}$

$E[(Y_n - X)^2] = E[Y_n^2] - E[2Y_nX]+E[X^2]= \frac{(n-1)^2}{n^2}E[X^2] - \frac{2(n-1)}{n}E[X^2] + E[X^2] = $

$=\frac{(n-1)^2}{3n^2} -\frac{2(n-1)}{3n} + \frac{1}{3} = \frac{n^2-2n+1-2n^2+2n+n^2}{3n^2} = \frac{+1}{3n^2} $

So: $ \Bbb D^2[Y_n-X] = \frac{1}{3n^2} - \frac{1}{4n^2} = \frac{1}{12n^2}$

And we're done, since:

$$$\Bbb P(|Y_n-X| > \epsilon) \leq \frac{D^2(Y_n-X)}{\epsilon^2} = \frac{1}{12n^2\epsilon^2} \to 0, as \ n \to \infty $$

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  • $\begingroup$ I didn't know before that I can manipulate and calculate $E[Y_n - X]$ instead. Does this trick always hold though? $\endgroup$ Feb 21, 2019 at 1:46
  • $\begingroup$ U mean the "trick" $E[Y_n - X] = E[Y_n] - E[X] $? If so, then answer is yes, expected value is linear. $\endgroup$
    – Presage
    Feb 21, 2019 at 1:49
  • $\begingroup$ I encountered some problems that simplifies the terms inside the modulus and then express it in the form of CDF. Now I know it is not applicable to all cases, especially when you are given $Y$ in terms of function of $X$ $\endgroup$ Feb 21, 2019 at 1:57
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It may help to write $|Y_n - X| = |X|/n$, and then use Markov or Chebychev.


More detail:

$|Y_n - X| = |\frac{n-1}{n} X - X| = |-X/n| = X/n$ since $X$ is nonnegative. Thus $$P(|Y_n - X| > \epsilon) = P(X/n > \epsilon) = P(X > n \epsilon).$$ You can use Markov's inequality: $$P(X > n \epsilon) = \frac{E[X]}{n\epsilon} \to 0,$$ or even Chebychev's inequality: $$P(X > n \epsilon) = P(X - \frac{1}{2} > n\epsilon - \frac{1}{2}) = \frac{\text{Var}(X)}{(n\epsilon - \frac{1}{2})^2} \to 0.$$

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  • $\begingroup$ How do you figure out $\frac{|X|}{n}$ ? $\endgroup$ Feb 21, 2019 at 1:22
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    $\begingroup$ @dembrownies $|Y_n - X| = |\frac{n-1}{n} X - X| = |-\frac{1}{n} X|$? $\endgroup$
    – angryavian
    Feb 21, 2019 at 1:25

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