If X is a Poisson variate such that `P(X=2)=9P(X=4)+90P(X=6)`, then mean of X is a) `1` b) ` 2` c) `1/2` d)`3/2`

Answer

To solve the problem, we need to find the mean of a Poisson random variable \( X \) given the equation:

\[ P(X=2) = 9P(X=4) + 90P(X=6) \]

### Step 1: Write the Poisson Probability Mass Function (PMF)

The PMF for a Poisson random variable \( X \) with mean \( \lambda \) is given by:

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

### Step 2: Substitute the PMF into the given equation

Using the PMF, we can express \( P(X=2) \), \( P(X=4) \), and \( P(X=6) \):

\[ P(X=2) = \frac{e^{-\lambda} \lambda^2}{2!} \]
\[ P(X=4) = \frac{e^{-\lambda} \lambda^4}{4!} \]
\[ P(X=6) = \frac{e^{-\lambda} \lambda^6}{6!} \]

Substituting these into the equation gives:

\[ \frac{e^{-\lambda} \lambda^2}{2} = 9 \cdot \frac{e^{-\lambda} \lambda^4}{24} + 90 \cdot \frac{e^{-\lambda} \lambda^6}{720} \]

### Step 3: Simplify the equation

We can simplify the right-hand side:

\[ 9 \cdot \frac{e^{-\lambda} \lambda^4}{24} = \frac{3}{8} e^{-\lambda} \lambda^4 \]
\[ 90 \cdot \frac{e^{-\lambda} \lambda^6}{720} = \frac{1}{8} e^{-\lambda} \lambda^6 \]

Now substituting these back into the equation:

\[ \frac{e^{-\lambda} \lambda^2}{2} = \frac{3}{8} e^{-\lambda} \lambda^4 + \frac{1}{8} e^{-\lambda} \lambda^6 \]

### Step 4: Cancel out \( e^{-\lambda} \)

Since \( e^{-\lambda} \) is common on both sides, we can cancel it out (assuming \( e^{-\lambda} \neq 0 \)):

\[ \frac{\lambda^2}{2} = \frac{3}{8} \lambda^4 + \frac{1}{8} \lambda^6 \]

### Step 5: Multiply through by 8 to eliminate fractions

Multiply the entire equation by 8:

\[ 4\lambda^2 = 3\lambda^4 + \lambda^6 \]

### Step 6: Rearrange the equation

Rearranging gives:

\[ \lambda^6 + 3\lambda^4 - 4\lambda^2 = 0 \]

### Step 7: Factor out \( \lambda^2 \)

Factoring out \( \lambda^2 \):

\[ \lambda^2(\lambda^4 + 3\lambda^2 - 4) = 0 \]

This gives us two cases:

1. \( \lambda^2 = 0 \) (which implies \( \lambda = 0 \), not valid for a Poisson distribution)
2. \( \lambda^4 + 3\lambda^2 - 4 = 0 \)

### Step 8: Let \( y = \lambda^2 \)

Let \( y = \lambda^2 \):

\[ y^2 + 3y - 4 = 0 \]

### Step 9: Solve the quadratic equation

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \]

Calculating the roots:

1. \( y = \frac{2}{2} = 1 \)
2. \( y = \frac{-8}{2} = -4 \) (not valid since \( y \) must be non-negative)

Thus, \( y = 1 \) implies \( \lambda^2 = 1 \) or \( \lambda = 1 \).

### Conclusion: Mean of \( X \)

The mean of \( X \) is \( \lambda = 1 \).

Thus, the answer is:

**Mean of \( X \) is \( 1 \)** (Option a).