Exam Solution: Symmetric i.i.d. Random Variables
Understanding the Core Condition
The problem asks about properties of independent identically distributed (i.i.d.) random variables $X_1, X_2, \cdots$ with a common density $f$, given that the density is symmetric, meaning $f(x) = f(-x)$ for all $x$. This symmetry implies $X_i$ follows the same distribution as $-X_i$ (denoted $X_i \stackrel{d}{=} -X_i$).
Analysis of Options
We evaluate each statement based on the implications of the symmetric density $f(x) = f(-x)$.
- Option 1 & 2 (Convergence): These statements concern the convergence of the sample mean $\frac{1}{n}\sum_{i=1}^n X_i$ to 0. Such convergence (in probability or almost surely) is guaranteed by the Law of Large Numbers, but only if the expectation $E[X]$ exists. The symmetry condition $f(x)=f(-x)$ implies $E[X]=0$ only if $E[X]$ exists (i.e., if $E[|X|] < \infty$). Since the existence of $E[X]$ is not guaranteed (e.g., Cauchy distribution), these options are not always correct.
- Option 3 (CLT Behavior): This statement involves the limiting distribution related to the Central Limit Theorem (CLT). CLT requires the existence of the variance, $Var(X)$, which in turn requires the existence of the mean, $E[X]$. As these are not guaranteed by the symmetry condition alone, this option is not universally correct.
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Option 4 (Distribution Equivalence): This statement asserts that $S_n = \sum_{i=1}^n X_i$ has the same distribution as $T_n = \sum_{i=1}^n (-1)^i X_i$. This property holds true based solely on the symmetry $f(x)=f(-x)$ and independence.
- Characteristic Functions: Let $\phi(t) = E[e^{itX}]$ be the characteristic function of $X$. Symmetry ($f(x)=f(-x)$) implies $\phi(t) = \phi(-t)$.
- The characteristic function of $S_n = \sum_{i=1}^n X_i$ is $\phi_{S_n}(t) = (\phi(t))^n$.
- The characteristic function of $T_n = \sum_{i=1}^n (-1)^i X_i$ is $\phi_{T_n}(t) = E[e^{it \sum (-1)^i X_i}]$. Due to independence, this equals $\prod_{i=1}^n E[e^{it(-1)^i X_i}]$.
- For each term, $E[e^{it(-1)^i X_i}] = \phi(t)$ because $\phi(-t)=\phi(t)$. Thus, $\phi_{T_n}(t) = (\phi(t))^n$.
- Since $\phi_{S_n}(t) = \phi_{T_n}(t)$, $S_n$ and $T_n$ have the same distribution.
Final Conclusion
Statement 4 is the only one that is guaranteed to be correct under the given conditions ($X_i$ are i.i.d. with symmetric density $f(x)=f(-x)$), without needing further assumptions about the existence of $E[X]$ or $Var(X)$.
Important Questions from Random Variables
Let $X_1$ and $X_2$ be a random sample from Uniform$[0, \theta]$ distribution, where $\theta > 0$. For testing the hypothesis
$H_0: \theta = 1$ against $H_1: \theta = 2$,
consider a test which rejects $H_0$ if $X_1 + X_2 > \frac{4}{5}$. Then, the probability of type-I error is- Let $\{Y_n: n \ge 1\}$ be a sequence of independent and identically distributed random variables, where $Y_1 \sim \text{Bernoulli}(\frac{1}{2})$. Define $Z = \sum_{n=1}^\infty \frac{4Y_n}{5^n}$. Then, which of the following statements is true?
Let $X$ be a single sample from an absolutely continuous distribution with probability density function
$f(x|\theta) = \begin{cases} \frac{2}{\theta^2}(\theta - x), & \text{if } 0 < x < \theta \\ 0, & \text{otherwise,} \end{cases}$
where $\theta > 0$ is unknown. Which of the following intervals is a $95\%$ confidence interval for $\theta$?- Let $X$ and $Y$ be two independent random variables such that the moment generating func-tions of $X$ and $Y$ are
$M_X(t) = e^{3(e^t-1)}, t \in R$,
and
$M_Y(t) = (\frac{1}{2}e^{-3t} + \frac{1}{2}e^{3t})^2, t \in R$,
respectively. Then which of the following statements are true? Let $X_1, X_2, ..., X_7$ be a random sample drawn from a continuous distribution with unknown unique median $M$. The null hypothesis $H_0: M = 2$ is tested against the alternative $H_1: M \gt 2$ at level of significance 0.05 using the right-tailed test based on the Sign test statistic $K$, which is the number of observations in the sample greater than 2. If the observed sample is $-3, -6, 1, 9, 4, 10, 12$, which of the following statements are true?
