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