$P(Z \ge \frac{3}{5}) = 0.5, P(Z = \frac{4}{25}) = 0$
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 $X$, $Y$, and $Z$ be independent Normal random variables with means $-1$, $0$, and $1$, respectively, and variances $1$, $1$, and $3$, respectively. Which of the following random variables has a Cauchy distribution with location parameter $0$ and scale parameter $1$?
- Consider a finite population of size $N = 100$. Let $T_1$ be the sample mean of a study variable based on a sample of size $n$ ($1 < n < N$) under simple random sampling with replacement scheme. Let $T_2$ be the sample mean of the same study variable based on a sample of size $n$ under simple random sampling without replacement scheme. If $Var(T_1) = 9Var(T_2)$, then the sample size $n$ equals
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 $X_1, X_2,..., X_{15}$ be a random sample from an Exponential distribution with the probability density function
$f(x) = \begin{cases} \frac{1}{\sigma} \exp \left(-\frac{x}{\sigma}\right) & \text{if } x > 0, \\ 0, & \text{elsewhere,} \end{cases}$
where the unknown parameter $\sigma$ is positive. Let $\bar{X} = \frac{1}{15}\sum_{i=1}^{15} X_i$. Suppose that $\phi$ denotes the likelihood ratio test for testing $H_0: \sigma \le 1$ against $H_1 : \sigma > 1$ at level $\alpha = 0.1$. It is given that $\chi^2_{15,0.1} = 22.307$, $\chi^2_{15,0.9} = 8.547$, $\chi^2_{30,0.1} = 40.256$, $\chi^2_{30,0.9} = 20.599$, where $P(W > \chi^2_{n,\alpha}) = \alpha$ and $W \sim \chi^2_n$. Then which of the following statements are true?
