If X is a Poisson variate such that `P(X=0)=P(X=1)`, then the parameter `lambda=`

Options

A. 1
B. 2
C. 3
D. `1//2`

Answer

To solve the problem, we need to find the parameter \(\lambda\) for a Poisson random variable \(X\) such that \(P(X=0) = P(X=1)\).

### Step-by-Step Solution:

1. **Write the Poisson Probability Mass Function (PMF)**:
The PMF for a Poisson random variable \(X\) with parameter \(\lambda\) is given by:
\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
where \(k\) is a non-negative integer.

2. **Set Up the Equation**:
We know that \(P(X=0) = P(X=1)\). Thus, we can write:
\[ P(X=0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-\lambda} \]
and
\[ P(X=1) = \frac{e^{-\lambda} \lambda^1}{1!} = e^{-\lambda} \lambda \]
Therefore, we have:
\[ e^{-\lambda} = e^{-\lambda} \lambda \]

3. **Simplify the Equation**:
Since \(e^{-\lambda}\) is common on both sides, we can divide both sides by \(e^{-\lambda}\) (noting that \(e^{-\lambda} \neq 0\)):
\[ 1 = \lambda \]

4. **Conclusion**:
Thus, we find that:
\[ \lambda = 1 \]

### Final Answer:
The parameter \(\lambda\) is equal to \(1\).

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