For a poisson variate X, P(X = 2) = 3P(X = 3) find variance of X.

Answer

To solve the problem, we need to use the properties of the Poisson distribution. The given information states that for a Poisson variate \( X \), \( P(X = 2) = 3P(X = 3) \). We will use the formula for the Poisson probability mass function (PMF):

\[ P(X = n) = \frac{\lambda^n e^{-\lambda}}{n!} \]

where \( \lambda \) is the mean (and also the variance) of the Poisson distribution.

### Step-by-step Solution:

1. **Write the expressions for \( P(X = 2) \) and \( P(X = 3) \)**:
\[ P(X = 2) = \frac{\lambda^2 e^{-\lambda}}{2!} \]
\[ P(X = 3) = \frac{\lambda^3 e^{-\lambda}}{3!} \]

2. **Set up the equation based on the given information**:
According to the problem, we have:
\[ P(X = 2) = 3P(X = 3) \]
Substituting the expressions from step 1:
\[ \frac{\lambda^2 e^{-\lambda}}{2!} = 3 \cdot \frac{\lambda^3 e^{-\lambda}}{3!} \]

3. **Simplify the equation**:
The factorials are \( 2! = 2 \) and \( 3! = 6 \), so we rewrite the equation:
\[ \frac{\lambda^2 e^{-\lambda}}{2} = 3 \cdot \frac{\lambda^3 e^{-\lambda}}{6} \]
This simplifies to:
\[ \frac{\lambda^2 e^{-\lambda}}{2} = \frac{3\lambda^3 e^{-\lambda}}{6} \]
Multiplying both sides by 6 to eliminate the fraction:
\[ 3\lambda^2 e^{-\lambda} = 3\lambda^3 e^{-\lambda} \]

4. **Cancel \( 3e^{-\lambda} \) from both sides** (assuming \( e^{-\lambda} \neq 0 \)):
\[ \lambda^2 = \lambda^3 \]

5. **Rearrange the equation**:
\[ \lambda^3 - \lambda^2 = 0 \]
Factor out \( \lambda^2 \):
\[ \lambda^2(\lambda - 1) = 0 \]

6. **Solve for \( \lambda \)**:
This gives us two solutions:
\[ \lambda^2 = 0 \quad \text{or} \quad \lambda - 1 = 0 \]
Thus, \( \lambda = 0 \) or \( \lambda = 1 \). Since \( \lambda = 0 \) does not make sense in the context of a Poisson distribution, we have:
\[ \lambda = 1 \]

7. **Find the variance**:
Since the variance of a Poisson distribution is equal to \( \lambda \), we have:
\[ \text{Variance} = \lambda = 1 \]

### Final Answer:
The variance of \( X \) is \( 1 \).