$P(X=1) = 4P (X=2)$
The variance of X is
Poisson Variance Calculation Explained
This problem asks us to find the variance of a random variable X that follows a Poisson distribution. We are given a specific relationship between two probabilities: $P(X=1) = 4P(X=2)$.
Understanding the Poisson Distribution
The Poisson distribution is a discrete probability distribution used to model the number of events occurring within a fixed interval of time or space. The probability mass function (PMF) for a Poisson random variable X with mean $\lambda$ (lambda) is given by:
$P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}$where:
- $k$ is the number of occurrences (a non-negative integer: 0, 1, 2, ...).
- $\lambda$ is the average rate of occurrences (mean, $\lambda > 0$).
- $e$ is the base of the natural logarithm (approximately 2.71828).
- $k!$ is the factorial of $k$.
A key property of the Poisson distribution is that its variance is equal to its mean:
$Var(X) = \lambda$Our goal is to use the given probability condition to find the value of $\lambda$, and that value will be the variance.
Applying the Probability Condition
First, let's write out the expressions for $P(X=1)$ and $P(X=2)$ using the PMF:
- For $k=1$: $P(X=1) = \frac{e^{-\lambda}\lambda^1}{1!} = \frac{e^{-\lambda}\lambda}{1} = e^{-\lambda}\lambda$
- For $k=2$: $P(X=2) = \frac{e^{-\lambda}\lambda^2}{2!} = \frac{e^{-\lambda}\lambda^2}{2} $
Now, we substitute these into the given equation $P(X=1) = 4P(X=2)$:
$e^{-\lambda}\lambda = 4 \times \left( \frac{e^{-\lambda}\lambda^2}{2} \right)$Solving for the Parameter $\lambda$
Let's simplify the equation to solve for $\lambda$:
- Cancel out $e^{-\lambda}$ from both sides (since $e^{-\lambda}$ is always positive and never zero): $\lambda = 4 \times \left( \frac{\lambda^2}{2} \right)$
- Simplify the right side: $\lambda = 2\lambda^2$
- Rearrange the equation to solve for $\lambda$: $2\lambda^2 - \lambda = 0$
- Factor out $\lambda$: $\lambda(2\lambda - 1) = 0$
- This equation gives two possible solutions for $\lambda$: $\lambda = 0$ or $2\lambda - 1 = 0$.
If $\lambda = 0$, the distribution is degenerate ($P(X=0)=1$), and $P(X=1)=0$, $P(X=2)=0$. The condition $0 = 4 \times 0$ holds, but the variance would be 0. Since the options include non-zero values, we consider the other solution.
Solving $2\lambda - 1 = 0$ gives:
$2\lambda = 1$ $\lambda = \frac{1}{2}$This is a valid parameter for a Poisson distribution ($\lambda > 0$).
Determining the Variance
As mentioned earlier, for a Poisson distribution, the variance is equal to the parameter $\lambda$. Since we found $\lambda = \frac{1}{2}$, the variance of X is:
$Var(X) = \lambda = \frac{1}{2}$Conclusion
The variance of the random variable X is $\frac{1}{2}$.
Important Questions from Probability Distribution
If the mean and variance of a binomial distribution are 5 and 4, respectively, then the value of n is:
For the distribution with unknown θ
\(f(x,\theta ) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{\theta };0 \le x \le \theta }\\ {0;elsewhere} \end{array}} \right.\)
We set the testing of hypothesis H 0 ∶ θ = 1 vs H 1 ∶ θ = 2. When the critical region X ≥ 0.4, the value of probability of type-II error is:
For the cumulative distribution function \(F(x) = \left\{ {\begin{array}{*{20}{c}} {0;x < - 1}\\ {\frac{1}{2}{{(x + 1)}^2}; - 1 \le x < 0}\\ {1 - \frac{{{{(1 - x)}^2}}}{2};0 \le x < 1}\\ {1.1 < x < \infty } \end{array}} \right.\)
the upper quartile point is
Let the joint probability density function of \( (X, Y) \) be
\[f(x, y) = \begin{cases} 6xy^2 & \text{if } 0 < x < 1, 0 < y < 1 \\ 0, & \text{otherwise} \end{cases}\]
Then \( P\left(\frac{1}{2} < X < \frac{3}{4}\right) \) is:
Let X and Y have the joint p.m.f. f(x, y) = x + y / 21, where x = 1, 2, 3 and y = 1, 2. The marginal p.m.f. of X is:
