How Do Probability Formulas for Bayes Theorem and Exponential Distribution Work?

  • Topic:
  • Thread starter Longines
  • Start date
  • Tags Tags
    Parameters
2 replies · 2K views · last reply
Longines
Messages
9
Reaction score
0
Hey guys,

I don't understand how this question works... I don't understand the answers either. Could someone take me through this step-by-step?

See attached image:
 
Attachments
  • Typeset worked solution: part (a) derives P(k-1 < X < k) = e^(-(k-1)λ) - e^(-kλ) = (1 - e^(-λ))(e^(-λ))^(k-1), so G is geometric with p = 1 - e^(-λ); part (b) gives P(X > x+k | G > k) = P(X > x) = e^(-λx) by memorylessness.
    Screenshot 2014-09-20 20.20.08.png
    39.1 KB · Views: 132

Discussion

Longines said:
Hey guys,

I don't understand how this question works... I don't understand the answers either. Could someone take me through this step-by-step?

See attached image:

a) is...

$\displaystyle P \{ G = k\} = \int_{k-1}^{k} e^{- \lambda\ x}\ d x = e^{\lambda\ k} (e^{\lambda} - 1)\ (1)$

b) for the Bayes theorem is...

$\displaystyle P\{X > k + x| G > k \} = \frac{P \{ X > k + x \}}{P\{X>k \}} = \frac{e^{- \lambda\ (k + x)}}{e^{- \lambda\ k}} = e^{- \lambda\ x}\ (2) $

Kind regards

$\chi$ $\sigma$
 
chisigma said:
a) is...

$\displaystyle P \{ G = k\} = \int_{k-1}^{k} e^{- \lambda\ x}\ d x = e^{\lambda\ k} (e^{\lambda} - 1)\ (1)$

b) for the Bayes theorem is...

$\displaystyle P\{X > k + x| G > k \} = \frac{P \{ X > k + x \}}{P\{X>k \}} = \frac{e^{- \lambda\ (k + x)}}{e^{- \lambda\ k}} = e^{- \lambda\ x}\ (2) $

Kind regards

$\chi$ $\sigma$
Lol... once again, a simple step that I did not see.

Thank you