I am struggling with a small part of the proof of the law of large numbers.

I understand from Markov's inequality: $$P(X\ge t) \le \frac{E(X)}{t} $$ and therefore if $ X = (\bar{Y} - E(Y))^2$

$$P(|\bar{Y} - E(Y)|\ge t) \le \frac{\sigma^2}{nt^2} $$

However, what i don't quite understand is what happens when we switch the inequality on the left from $\ge t$ to $\le t$.

I understand the fact that if $P(X>t)=c$ then $P(X<t)=1-c$, however I am finding it hard to think about why the correct answer is:

$$P(|\bar{Y} - E(Y)|\le t) \ge 1-\frac{\sigma^2}{nt^2} $$

and not:

$$P(|\bar{Y} - E(Y)|\le t) \le 1-\frac{\sigma^2}{nt^2} $$

  • $\begingroup$ It's a typo. Your answer is correct and your argument proves it. $\endgroup$
    – Michael
    Feb 12, 2017 at 20:28
  • $\begingroup$ Sorry I haven't worded it correctly, but I am not clear why either of the two options is correct. I don't fully understand why the less than sign is changed to greater than. $\endgroup$ Feb 12, 2017 at 20:45
  • 2
    $\begingroup$ If you have $1-c \leq A$ then $-c \leq A-1$ then $ c\geq 1-A$. Does that help? For you $A=\sigma^2/n t^2$. $\endgroup$
    – Michael
    Feb 12, 2017 at 20:48
  • $\begingroup$ Regarding your first comment, I've changed the order of the equations so my original question makes sense. Thanks for your follow up comment, I was originally just converting the right-hand side to 1-c and being confused why the inequality doesn't match. I realise I should convert the left side to 1-P(...<t) then work forwards using your method to get the right answer! $\endgroup$ Feb 12, 2017 at 21:25
  • $\begingroup$ Since you appear to have arrived at an answer based on Michael's comments, please post it (with proper attribution of the ideas contributed by Michael, naturally) $\endgroup$
    – Glen_b
    Feb 13, 2017 at 2:23

1 Answer 1


So we know that:

$$P(|\bar{Y} - E(Y)|\ge t) = 1 - P(|\bar{Y} - E(Y)|\le t) $$

and from the question:

$$P(|\bar{Y} - E(Y)|\ge t) \le \frac{\sigma^2}{nt^2} $$


$$1 - P(|\bar{Y} - E(Y)|\le t) \le \frac{\sigma^2}{nt^2} $$

Equal to :

$$1 \le \frac{\sigma^2}{nt^2} + P(|\bar{Y} - E(Y)|\le t) $$ $$1 - \frac{\sigma^2}{nt^2} \le P(|\bar{Y} - E(Y)|\le t) $$

So as n approaches infinity:

$$1 \le P(|\bar{Y} - E(Y)|\le t) $$

Thanks to Michael for clearing up the problem in the comments.


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