approximate probability of geometric distribution using CLT
$begingroup$
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?
central-limit-theorem
$endgroup$
add a comment |
$begingroup$
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?
central-limit-theorem
$endgroup$
add a comment |
$begingroup$
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?
central-limit-theorem
$endgroup$
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?
central-limit-theorem
central-limit-theorem
edited Dec 10 '18 at 13:29
amWhy
1
1
asked Nov 27 '18 at 14:40
NourNour
234
234
add a comment |
add a comment |
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