Poisson Normal Approximation Solution
The question asks to identify expressions equal to the integral $\frac{1}{\sqrt{2\pi}} \int_2^\infty e^{-x^2/2} dx$, which represents $P(Z > 2)$ for a standard normal variable $Z$. We use the Normal Approximation to the Poisson distribution.
Normal Approximation to Poisson
For large $n$, a Poisson random variable $X_n$ with mean $\lambda_n = n^2$ can be approximated by a normal distribution with mean $\mu_n = n^2$ and variance $\sigma_n^2 = n^2$. The standard deviation is $\sigma_n = \sqrt{n^2} = n$. The standardized variable is $Z_n = \frac{X_n - \mu_n}{\sigma_n} = \frac{X_n - n^2}{n}$. As $n \to \infty$, $Z_n$ converges in distribution to the standard normal variable $Z$.
Option Analysis
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Option 1: $\lim_{n \to \infty} P\{X_n > (n+1)^2\}$
Standardizing the term:
$ P\left\{X_n > (n+1)^2\right\} = P\left\{\frac{X_n - n^2}{n} > \frac{(n+1)^2 - n^2}{n}\right\} $ $ \frac{(n+1)^2 - n^2}{n} = \frac{n^2 + 2n + 1 - n^2}{n} = \frac{2n + 1}{n} = 2 + \frac{1}{n} $Taking the limit:
$ \lim_{n \to \infty} P\left\{Z_n > 2 + \frac{1}{n}\right\} = P(Z > 2) $This matches the given integral $\frac{1}{\sqrt{2\pi}} \int_2^\infty e^{-x^2/2} dx$. Thus, Option 1 is correct.
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Option 2: $\lim_{n \to \infty} P\{X_n \le (n+1)^2\}$
Following the standardization from Option 1:
$ \lim_{n \to \infty} P\left\{X_n \le (n+1)^2\right\} = \lim_{n \to \infty} P\left\{Z_n \le 2 + \frac{1}{n}\right\} = P(Z \le 2) $This is equal to $1 - P(Z > 2)$, which does not match the integral.
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Option 3: $\lim_{n \to \infty} P\{X_n < (n-1)^2\}$
Standardizing the term:
$ P\left\{X_n < (n-1)^2\right\} = P\left\{\frac{X_n - n^2}{n} < \frac{(n-1)^2 - n^2}{n}\right\} $ $ \frac{(n-1)^2 - n^2}{n} = \frac{n^2 - 2n + 1 - n^2}{n} = \frac{-2n + 1}{n} = -2 + \frac{1}{n} $Taking the limit:
$ \lim_{n \to \infty} P\left\{Z_n < -2 + \frac{1}{n}\right\} = P(Z < -2) $Due to the symmetry of the standard normal distribution, $P(Z < -2) = P(Z > 2)$. This matches the given integral.
Thus, Option 3 is correct. -
Option 4: $\lim_{n \to \infty} P\{X_n < (n-2)^2\}$
Standardizing the term:
$ P\left\{X_n < (n-2)^2\right\} = P\left\{\frac{X_n - n^2}{n} < \frac{(n-2)^2 - n^2}{n}\right\} $ $ \frac{(n-2)^2 - n^2}{n} = \frac{n^2 - 4n + 4 - n^2}{n} = \frac{-4n + 4}{n} = -4 + \frac{4}{n} $Taking the limit:
$ \lim_{n \to \infty} P\left\{Z_n < -4 + \frac{4}{n}\right\} = P(Z < -4) $This does not match the integral $P(Z > 2)$.
Conclusion
The expressions equal to $\frac{1}{\sqrt{2\pi}} \int_2^\infty e^{-x^2/2} dx$ are found in Option 1 and Option 3.
Important Questions from Central Limit Theorems
- Let $X_1, X_2, . . .$ be a sequence of independent and identically distributed random variables with $E(X_1) = 0, E(X_1^2) = 1, E(X_1^3) = 0, E(X_1^4) = 3$. Let $S_n = \sum_{i=1}^n X_i, T_n = \sum_{i=1}^n X_i^2, U_n = \sum_{i=1}^n X_i^3$ and $V_n = \sum_{i=1}^n X_i^4$. Then, which of the following statements are true?
- Let $\{X_i; i \ge 1\}$ be a sequence of independent random variables each having a normal distribution with mean 2 and variance 5. Then which of the following are true
Let $\{X_n : n \ge 1\}$ be a sequence of independent and identically distributed random variables and the probability mass function of $X_1$ is the following;
$P(X_1 = 1) = P(X_1 = 3) = \frac{1}{2}.$ If $Y_n = X_1 + \cdots + X_n$,
then which of the following statements are correct?
Suppose $X_1, X_2, \dots$ are independent random variables. Assume that $X_1, X_3, \dots$ are identically distributed with mean $\mu_1$ and variance $\sigma_1^2$, while $X_2, X_4, \dots$ are identically distributed with mean $\mu_2$ variance $\sigma_2^2$. Let $S_n = X_1 + X_2 + \dots + X_n$. Then $\frac{S_n - a_n}{b_n}$ converges in distribution to $N(0,1)$ if
- Let $X_i$'s be independent random variables such that $X_i$'s are symmetric about 0 and $\text{Var}(X_i) = 2i-1$, for $i \ge 1$. Then,
$$\lim_{n\to\infty} P(X_1 + X_2 + \cdots + X_n > n \log n)$$
