We begin by analyzing the given problem involving independent Poisson random variables \(X\) and \(Y\) with means 4 and 2 respectively. We need to evaluate which of the given statements are true.
1. **Conditional Distribution**: The statement claims that the conditional distribution of \(X\) given \(X + Y = 3\) is a Binomial distribution. For Poisson random variables \(X\) and \(Y\), their joint distribution given \(X + Y = n\) follows a binomial distribution.
The distribution is \(X | (X + Y = n) \sim \text{Binomial}(n, \frac{\lambda_1}{\lambda_1 + \lambda_2})\), where \(\lambda_1\) and \(\lambda_2\) are the means of \(X\) and \(Y\).
Here, \(\lambda_1 = 4\) and \(\lambda_2 = 2\), so \(X| (X + Y = 3) \sim \text{Binomial}(3, \frac{4}{6}) = \text{Binomial}(3, \frac{2}{3})\).
Therefore, the statement that the conditional distribution is \(\text{Binomial}(3, \frac{1}{3})\) is false.
2. **Probability Calculation**: \(P(X \le 1 | X + Y = 3)\) should be computed to verify if it equals \(\frac{7}{27}\).
The pmf for a Binomial random variable is \(P(Z = k) = \binom{n}{k} p^k (1-p)^{n-k}\).
Here, \(P(X = 0 | X + Y = 3) = \binom{3}{0}(\frac{2}{3})^0(\frac{1}{3})^3 = \frac{1}{27}\)
\(P(X = 1 | X + Y = 3) = \binom{3}{1}(\frac{2}{3})^1(\frac{1}{3})^2 = \frac{6}{27} = \frac{2}{9}\)
Thus, \(P(X \le 1 | X+Y=3) = \frac{1}{27} + \frac{6}{27} = \frac{7}{27}\).
This matches the given statement, so it is true.
3. **Expected Value Calculation**: \(E(X | X+Y = 3)\) is calculated using the mean of the binomial distribution, which is \(n \cdot p\).
Here, \(E(X | X + Y = 3) = 3 \cdot \frac{2}{3} = 2\).
This matches the given statement, so it is true.
4. **Characteristic Function**: The characteristic function for a Poisson distribution with mean \(\lambda\) at a point \(t\) is given by \(e^{\lambda(e^{it} - 1)}\).
The sum of two independent Poisson random variables is another Poisson random variable with mean equal to the sum of their means, i.e., mean 6 here.
Thus, the characteristic function of \(X+Y\) at \(t = \pi\) is \(e^{6(e^{i\pi} - 1)} = e^{6(-2)} = e^{-12}\), which aligns with the statement.
Based on the analysis, the true statements are:
- \(P(X \le 1 | X+Y=3) = \frac{7}{27}\)
- \(E(X | X+Y = 3) = 2\)
- The value of the characteristic function of \(X + Y\) at the point \(t = \pi\) is \(e^{-12}\)
Important Questions from Discrete Probability
- A biased six-faced die is tossed once. Suppose that the probability of any prime number showing up is twice that of any non-prime number showing up. Then, the probability that an odd number will show up is
- Consider the M/M/1 queue in which customers arrive according to a Poisson process with rate $3$ and successive service times are independent exponential random variables having mean $\frac{1}{9}$. Let $P_n$ be the long run probability that there are exactly $n$ customers in the system. Then, which of the following statements are true?
- Let $X$ be a Binomial$(n, p)$ random variable, where $n \in \{5,6\}$ and $p\in \{\frac{1}{4}, \frac{3}{4}\}$. If $X = 3$ is observed, then the maximum likelihood estimate of $(n, p)$ is
- Suppose two fair dice are thrown independently at random. Let $X$ and $Y$ be the numbers on the upper face of the first die and that of the second die, respectively. Then which of the following statements are true?
- Suppose customers arrive in a shop according to a Poisson process with rate 4 per hour. The shop opens at 10:00 am. If it is given that the second customer arrives at 10:40 am, what is the probability that no customer arrived before 10:30 am?
