Question

In a Poisson distribution, \[ P(X=x+k)=\frac{\lambda^k}{f(x)}P(X=x), \] then \(f(x)=\)

• A. \(\displaystyle \frac{e^{-\lambda}\lambda^x}{x!}\)
• B. \(\displaystyle \frac{e^{-\lambda}\lambda^{x+k}}{(x+k)!}\)
• C. \(\displaystyle (x+k)(x+k-1)\cdots(x+1)\)
• D. \(\displaystyle \frac{(x+k)(x+k-1)\cdots(x+1)}{x!}\)

Hint:

For a Poisson distribution, \[ P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!}. \] When taking ratios of probabilities, the common factor \(e^{-\lambda}\) always cancels, simplifying the calculation considerably.

Answer 1

The Correct Option is C

Concept: The probability mass function of a Poisson random variable with parameter \(\lambda\) is \[ P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!}, \qquad r=0,1,2,\ldots \] Using this formula, the ratio of two probabilities can be simplified to obtain the required expression.

Step 1: Write the probabilities. For a Poisson distribution, \[ P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}, \] and \[ P(X=x+k)=\frac{e^{-\lambda}\lambda^{x+k}}{(x+k)!}. \]

Step 2: Find the ratio. \[ \frac{P(X=x+k)}{P(X=x)} = \frac{e^{-\lambda}\lambda^{x+k}}{(x+k)!} \cdot \frac{x!}{e^{-\lambda}\lambda^x} = \frac{\lambda^k\,x!}{(x+k)!}. \] Since \[ P(X=x+k)=\frac{\lambda^k}{f(x)}P(X=x), \] comparing both expressions, \[ \frac{\lambda^k}{f(x)} = \frac{\lambda^k x!}{(x+k)!}. \] Hence, \[ f(x) = \frac{(x+k)!}{x!}. \] Now, \[ \frac{(x+k)!}{x!} = (x+k)(x+k-1)\cdots(x+1). \] Therefore, \[ \boxed{f(x)=(x+k)(x+k-1)\cdots(x+1).} \] Thus, the correct option is \[ \boxed{(C)\;(x+k)(x+k-1)\cdots(x+1).} \]