# Convergence in probability using normal distribution

I need help in solving this question:

Let $$Y_1$$, $$Y_2$$, $$...$$ be $$i.i.d.$$ with distribution $$N(2, 5)$$. Prove that for some $$n$$, we have $$P(Y_1 + Y_2 +···+ Y_n > n) > 0.999$$

I know from this, I need to find the new value of $$\bar{Y}$$ first, which is $$N\left(2, \frac{5}{n}\right)$$

How do I proceed from here?

• I don´t think that this is true. Have you calculated the case if $n=1$? – callculus Feb 20 at 21:38
• The answer should be $n \gt 5000$ – dembrownies Feb 21 at 1:17

## 2 Answers

By Strong Law of Large Numbers $$\frac {Y_1+Y_2+\cdots+Y_n} n \to 2$$ almost surely . This implies $$P(Y_1+Y_2+\cdots+Y_n >n) \to 1$$. Note that the exact distribution of $$Y_i$$ is not required for this and even the variance is not required. CLT is an overkill for this question.

• How do you know the probability converges to 1? – dembrownies Feb 21 at 1:18
• $P( Y_1+...+Y_n \leq n) \leq P( |\frac {Y_1+...+Y_n} n -2| \geq 1) \to 0$. – Kavi Rama Murthy Feb 21 at 5:26

You know $$\bar{Y} \sim N(2, 5/n)$$. Thus $$Z:=\sqrt{\frac{n}{5}}(\bar{Y} - 2) \sim N(0,1)$$. Further, $$P(Y_1 + \cdots + Y_n > n) = P(\bar{Y} > 1) = P(Z > -\sqrt{n/5}).$$ As $$n \to \infty$$, this quantity tends to $$1$$.

As noted in another answer, there are more direct methods (under even more relaxed assumptions) to show that there exists an $$n$$ for which the inequality holds. The virtue of this method (which uses knowledge of the distribution of each $$Y_i$$) is that one can compute for which specific values of $$n$$ the inequality holds.