How to derive the cdf of given function

Question

Let $$Y = \frac{X_1+X_2+\cdots+X_{100}}{100}$$

I understand that the Central Limit Theorem says you can approximate Y.

But by what random variable?

And how can I write the cumulative distribution function of this random variable?

I am confused on how to derive the CDF. Do I need to find the PDF first?

Answer 1

The CLT claims that the sum of a large number of $i.i.d.$ random variables are approximately normal distributed. One thumb of rule is that large number is more than $30$.

Let´s say $Y=\sum_{i=1}^{100} X_i$. Then $Y$ is approximately distributed as $Y\sim\mathcal N(100\cdot \mu_x, 100\cdot \sigma_x^2)$

$\mu_x is $the identical expected value for every $X_i$

$\sigma_x$ is the identical standard deviation for every $X_i$

Therefore $P(Y\leq y)\approx \Phi\left( \frac{y-100\cdot \mu_x}{\sqrt{100\cdot \sigma_x^2}} \right)$

where $\Phi\left( z \right)$ is the cdf of the standard normal distribution.

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Now we calculate the expected value and the variance of $\frac1{100}Y=\overline Y$

• $\mathbb E\left(\frac1{100}Y\right)=\frac1{100}\mathbb E\left(Y\right)$
• $Var\left(\frac1{100}Y\right)=\left(\frac1{100}\right)^2Var\left(Y\right)$

Thus $\frac1{100} Y=\overline Y$ is approximately distributed as $\overline Y\sim \mathcal N( \mu_x, \frac1{100}\sigma_x^2)$.

Finally we have

$$P(\overline Y\leq y)\approx \Phi\left( \frac{y- \mu_x}{\sqrt{\frac1{100}\cdot \sigma_x^2}} \right)=\Phi\left( \frac{y- \mu_x}{\frac1{10}\cdot \sigma_x} \right)$$