$P(Z \ge \frac{3}{5}) = 0.5, P(Z = \frac{4}{25}) = 0$
To solve this problem, we need to analyze the behavior of the random variable \(Z\) defined in the question. The sequence \(\{Y_n: n \ge 1\}\) is composed of independent and identically distributed Bernoulli random variables, each with a success probability of \(\frac{1}{2}\). For each \(n\), \(Y_n\) takes the value 1 with probability \(\frac{1}{2}\) and 0 with probability \(\frac{1}{2}\).
The random variable \(Z\) is defined as:
\(Z = \sum_{n=1}^\infty \frac{4Y_n}{5^n}\).
This series can be seen as a random infinite series with binary coefficients determined by \(Y_n\). Each \(Y_n\) contributes either \(0\) or \(\frac{4}{5^n}\) to the sum.
Calculating the Probability \(P(Z \ge \frac{3}{5})\)
To find \(P(Z \ge \frac{3}{5})\), we consider the contributions from the first few terms:
- The first term, \(\frac{4Y_1}{5}\), can be either \(0\) or \(\frac{4}{5}\).
- For \(Z\) to be at least \(\frac{3}{5}\), at least one of these terms must contribute sufficiently to reach this threshold.
If the first term contributes \(\frac{4}{5}\) (i.e., \(Y_1 = 1\)), we already exceed \(\frac{3}{5}\). Since \(Y_1 \sim \text{Bernoulli}\left(\frac{1}{2}\right)\), this happens with probability \(\frac{1}{2}\).
Thus, \(P(Z \ge \frac{3}{5}) = 0.5\).
Calculating the Probability \(P(Z = \frac{4}{25})\)
For \(Z = \frac{4}{25}\), we require the first term to contribute \(0\) and the exact sum from subsequent terms to equal \(\frac{4}{25}\). This is not possible because:
- \(\frac{4}{25}\) requires \(\frac{4Y_2}{25} = \frac{4}{25}\), which implies \(Y_2 = 1\), contributing exactly \(\frac{4}{25}\).
- However, since \(Z\) is an infinite sum and other terms also contribute based on their Bernoulli \(Y_n\) outcomes, exact attainment of \(\frac{4}{25}\) becomes impossible as this requires precise matching at infinity, leading to a zero probability.
Therefore, \(P(Z = \frac{4}{25}) = 0\).
Conclusion
The correct choice is the option stating \(P(Z \ge \frac{3}{5}) = 0.5, P(Z = \frac{4}{25}) = 0\). Hence, it confirms that this option accurately describes the distribution of \(Z\) based on how the series behaves.
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 isLet $X$ 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?
- $X_1, X_2, \cdots$ 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?
