I have the following problem:
For $i≥1$, let $X_i∼G_1/2$ be distributed Geometrically with parameter 1/2. Define
$$Y_n=\frac{1}{\sqrt{n}}\sum_{i=1}^n (X_i-2)$$
Approximate $P(−1≤Y_n≤2)$ with large enough $n$. Hint, note that $Y_n$ is not "properly" normalized.
I tried further normalize $Y_n$ by $Z_n=\frac{nY_n}\sigma$ and use $n=30,$ but then I am getting large values when applying the same normalization to -1 and 2. Any idea how to solve this problem?
To normalize $Y_n$ you need to divide by $\sigma$. Let me explain:
Notice that because $X_i$ is distributed geometrically with parameter 1/2, we know that $\mu = \frac{1}{1/2}=2$ and $\sigma = \frac{\sqrt{1-p}}{p} = \frac{2}{\sqrt{2}}$. Furthermore, the normalization of the sum of all $X_i$ is $\frac{\sum_{i=1}^{n}X_i - \mu n}{\sigma \sqrt{n}} = \frac{\sum_{i=1}^{n}X_i - 2 n}{2\sqrt{n}/\sqrt{2}}$. this could further more be written as: $$ \frac{\sqrt{2}}{2}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(X_i - 2)\\ =\frac{1}{\sigma}\frac{1}{\sqrt{n}}\sum_{i=1}^{n}(X_i - 2) $$
Finally, you just have to use a z-table to get the probabilities such that $P(\frac{-1}{\sigma}\leq Y \leq \frac{2}{\sigma}) = 0.6818$.