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

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SUMMARY

This discussion focuses on the application of Bayes' Theorem and the Exponential Distribution in probability calculations. The first formula presented, $\displaystyle P \{ G = k\} = \int_{k-1}^{k} e^{- \lambda\ x}\ d x = e^{\lambda\ k} (e^{\lambda} - 1)$, illustrates the probability of a random variable G taking a specific value k. The second formula, $\displaystyle P\{X > k + x| G > k \} = e^{- \lambda\ x}$, demonstrates how to calculate conditional probabilities using Bayes' Theorem. Both formulas are essential for understanding probabilistic modeling in statistics.

PREREQUISITES
  • Understanding of Exponential Distribution
  • Familiarity with Bayes' Theorem
  • Basic calculus for integration
  • Knowledge of probability theory
NEXT STEPS
  • Study the properties of Exponential Distribution in-depth
  • Explore advanced applications of Bayes' Theorem in statistical inference
  • Learn integration techniques relevant to probability calculations
  • Investigate real-world examples of probabilistic modeling
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Statisticians, data scientists, and anyone involved in probabilistic modeling or statistical analysis will benefit from this discussion.

Longines
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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:
 

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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
 

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