Exponential Distribution Formula

The exponential distribution in probability is the probability distribution that describes the time between events in a Poisson process.

Probability Density Function

\(\large f(x \lambda ) = \left\{\begin{matrix} \lambda e^{-\lambda x} & x >= 0,\\ 0 & x < 0. \end{matrix} \right \}\)

Cumulative Distribution Function

\[\large F(x ; \lambda ) = \left\{\begin{matrix} 1 – e^{-\lambda x} & x>= 0,\\ 0 & x < 0. \end{matrix}\right.\]

where $\lambda > 0$ is called the rate of the distribution.

The mean of the Exponential ($\lambda$) Distribution is calculated using integration by parts as –

\[\large E(X) = \int_{0}^{\infty } x\lambda e^{-\lambda x} \; dx\]

\[\large = \lambda \left [ \frac{-x \; e^{-\lambda x}}{\lambda}|_{0}^{\infty } + \frac{1}{\lambda }\int_{0}^{\infty } e^{-\lambda x} dx \right ]\]

\[\large = \lambda \left [ 0 + \frac{1}{\lambda }\frac{-e^{-\lambda x}}{\lambda} |_{o}^{\infty }\right ]\]

\[\large = \lambda \frac{1}{\lambda ^{2} }\]

\[\large = \frac{1}{\lambda }\]

 


Practise This Question

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes5T3 . Then the ratio of mM is: