$M_X(t) = e^{3(e^t-1)}, t \in R$,
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?
Distribution of Random Variable X
The moment generating function (MGF) of X is given by \(M_X(t) = e^{3(e^t-1)}\). This is the MGF of a Poisson distribution with parameter \(\lambda = 3\). Therefore, X follows a Poisson(3) distribution.
- Mean of X: \(E(X) = \lambda = 3\)
- Variance of X: \(Var(X) = \lambda = 3\)
Properties of Random Variable Y
The MGF of Y is \(M_Y(t) = (\frac{1}{2}e^{-3t} + \frac{1}{2}e^{3t})^2\). This can be written as \(M_Y(t) = (\cosh(3t))^2\). This MGF corresponds to the sum of two independent and identically distributed random variables, \(Z_1\) and \(Z_2\), where each \(Z_i\) takes values -3 and 3 with equal probability (1/2).
- Distribution of \(Z_i\): \(P(Z_i = -3) = 1/2\), \(P(Z_i = 3) = 1/2\)
- Mean of \(Z_i\): \(E(Z_i) = (-3)(\frac{1}{2}) + (3)(\frac{1}{2}) = 0\)
- Variance of \(Z_i\): \(E(Z_i^2) = (-3)^2(\frac{1}{2}) + (3)^2(\frac{1}{2}) = 9(\frac{1}{2}) + 9(\frac{1}{2}) = 9\). So, \(Var(Z_i) = E(Z_i^2) - (E(Z_i))^2 = 9 - 0^2 = 9\).
- Mean of Y: \(Y = Z_1 + Z_2\), so \(E(Y) = E(Z_1) + E(Z_2) = 0 + 0 = 0\).
- Variance of Y: Since \(Z_1\) and \(Z_2\) are independent, \(Var(Y) = Var(Z_1) + Var(Z_2) = 9 + 9 = 18\).
- Probability distribution of Y:
- \(P(Y = -6) = P(Z_1=-3, Z_2=-3) = (1/2)(1/2) = 1/4\)
- \(P(Y = 0) = P(Z_1=-3, Z_2=3) + P(Z_1=3, Z_2=-3) = (1/2)(1/2) + (1/2)(1/2) = 1/4 + 1/4 = 1/2\)
- \(P(Y = 6) = P(Z_1=3, Z_2=3) = (1/2)(1/2) = 1/4\)
Evaluating Statement 1: \(P(XY = 0)\)
The event \(XY = 0\) occurs if \(X=0\) or \(Y=0\) (or both). It is easier to calculate the complement: \(P(XY = 0) = 1 - P(X \neq 0 \text{ and } Y \neq 0)\).
Since X and Y are independent, \(P(X \neq 0 \text{ and } Y \neq 0) = P(X \neq 0) \times P(Y \neq 0)\).
- For X (Poisson(3)): \(P(X=0) = \frac{e^{-3}3^0}{0!} = e^{-3}\).
- Therefore, \(P(X \neq 0) = 1 - P(X=0) = 1 - e^{-3}\).
- For Y: \(P(Y=0) = 1/2\).
- Therefore, \(P(Y \neq 0) = 1 - P(Y=0) = 1 - 1/2 = 1/2\).
- \(P(X \neq 0 \text{ and } Y \neq 0) = (1 - e^{-3}) \times (1/2) = \frac{1}{2} - \frac{1}{2}e^{-3}\).
- \(P(XY = 0) = 1 - (\frac{1}{2} - \frac{1}{2}e^{-3}) = 1 - \frac{1}{2} + \frac{1}{2}e^{-3} = \frac{1}{2} + \frac{1}{2}e^{-3} = \frac{1+e^{-3}}{2}\).
Statement 1 is true.
Evaluating Statement 2: \(E(X + Y)\)
Using the linearity of expectation:
\(E(X + Y) = E(X) + E(Y)\)
\(E(X + Y) = 3 + 0 = 3\).
Statement 2 is true.
Evaluating Statement 3: \(Var(X + Y)\)
Since X and Y are independent random variables:
\(Var(X + Y) = Var(X) + Var(Y)\)
\(Var(X + Y) = 3 + 18 = 21\).
Statement 3 is true.
Evaluating Statement 4: \(Cov(X + Y, X - Y)\)
Using the properties of covariance:
\(Cov(X + Y, X - Y) = Cov(X, X) - Cov(X, Y) + Cov(Y, X) - Cov(Y, Y)\)
\(Cov(X + Y, X - Y) = Var(X) - Cov(X, Y) + Cov(Y, X) - Var(Y)\)
Since X and Y are independent, \(Cov(X, Y) = 0\) and \(Cov(Y, X) = 0\).
\(Cov(X + Y, X - Y) = Var(X) - Var(Y) = 3 - 18 = -15\).
The statement claims the covariance is 0, but it is -15.
Statement 4 is false.
Conclusion
The true statements are 1, 2, and 3.
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 $X_1, X_2, ..., X_7$ 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?
- $X_1, X_2, \cdots$ are independent identically distributed random variables having common density $f$. Assume $f(x) = f(-x)$ for all $x \in \mathbb{R}$. Which of the following statements is correct?
