A continuous random variable whose probability density function is given for some k > 0 by
F(x)= \[ke^{-kx}\] if x > 0 or equal to 0
and f(x)=0 if x< 0 is said to be exponential random variable with parameter k
Boundary condition k > 0
Mean of exponential distribution= \[\frac{1}{k}\]
Variance of exponential distribution=\[\frac{1}{k^2}\]
Results: for k=2 Mean =0.5 and variance =0.25
Example A continuous random variable has PDF of C\[e^{\frac{-x}{5}}\] , x > 0 find mean and variance of the given distribution.
solution
Since it is of the form of exponential distribution hence
Mean of exponential distribution= \[\frac{1}{C}\]
Variance of exponential distribution=\[\frac{1}{C^2}\]
for any PDF\[\int_{-\infty }^{\infty }Ce^{\frac{-x}{5}}dx =1\]
hence \[\int_{0 }^{\infty }Ce^{\frac{-x}{5}}dx =1\]
sovling this integration we get C=0.2
hence mean =5
and vraince =25
