Question

If a discrete random variable \(X\) has the probability distribution \[ P(X = x) = k \frac{2^{2x+1}}{(2x+1)!}, \quad x=0,1,2,\ldots, \] then find \(k\).

• A. \(\sinh 2\)
• B. \(\sec 2\)
• C. \(\cosech 2\)
• D. \(\cosh 2\)

Hint:

Use normalization of probability distribution and standard series expansions.

Answer 1

The Correct Option is C

Step 1: Use normalization condition
\[ \sum_{x=0}^\infty P(X=x) = 1 \implies k \sum_{x=0}^\infty \frac{2^{2x+1}}{(2x+1)!} = 1 \] Step 2: Recognize series
\[ \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!} = \sinh x \] So, \[ \sum_{x=0}^\infty \frac{2^{2x+1}}{(2x+1)!} = \sinh 2 \] Step 3: Find \(k\)
\[ k \sinh 2 = 1 \implies k = \frac{1}{\sinh 2} = \cosech 2 \]