Write an expected value and mathematical dispersion for exponential distribution.

Question

All I know is PDF for the exponential distribution $f(x)=\lambda e^{-\lambda x }, \text{if} x\ge 0 \text{ and } 0 \text{, if x < 0}$. Then how from there do I know the probability? If I guess and $\lambda$ is the amount of times the experiment is run then the expected value would be equal to $1$, but I really doubt it.

Answer 1

The expected value is computed by $$ \int_0^\infty (1-F(x))\ \mathsf dx = \int_0^\infty e^{-\lambda x} = \frac1\lambda. $$ Adding on to George Dewhirst's comment, the dispersion is given by the variance divided by the expected value. We compute using integration by parts $$ \mathbb E[X^2] = \int_0^\infty x^2 \lambda e^{-\lambda x} = \frac2{\lambda^2}. $$ Hence the variance of $X$ is given by $$\mathrm{Var}(X) = \mathbb E[X^2] = \mathbb E[X]2 = \frac2{\lambda^2} - \left(\frac1\lambda\right)^2 = \frac1{\lambda^2}.$$ It follows that the dispersion of $X$ is given by $$ \frac{\mathrm{Var}(X)}{\mathbb E[X]} = \frac{\frac1{\lambda^2}}{\frac1\lambda} = \frac1\lambda. $$