Question

If X is an exponential random variable with parameter λ, and c > 0, show that cX is exponential with parameter λ/c

Short answer

With differentiating, we have that,

$\frac{d}{dx}{F}_{cX}\left(x\right)=\frac{d}{dx}{F}_X\left(\frac{x}{c}\right)\cdot \frac{1}{c}=\frac{1}{c}{f}_X\left(\frac{x}{c}\right)=\frac{1}{c}\lambda e^{-\lambda \frac{x}{c}}$

$\text{Finally, we have obtained that for}x\ge 0\text{we have that}$

$f_{c}X\left(x\right)=\frac{\lambda }{c}{e}^{-\frac{\lambda }{c}x}$

$\text{and otherwise it is equal to zero. Hence, we have proved that}cX~\mathrm{Expo}\left(\frac{\lambda }{c}\right)$

Step-by-step solution

Given Information

If X is an exponential random variable with parameter λ, and c > 0, show that cX is exponential with parameter λ/c

Explanation

$\text{Since we know that}P(X\ge 0)=1\text{, we have that}$

$P(cX\ge 0)=P(X\ge 0)=1$

$\text{so}cX\text{is also non- negative random variable. The distribution of}cX\text{is}$

$F_{cX}\left(x\right)=P(cX\le x)=P(X\le \frac{x}{c})=F_X\left(\frac{x}{c}\right)$

So, with differentiating, we have that,

$\frac{d}{dx}{F}_{cX}\left(x\right)=\frac{d}{dx}{F}_X\left(\frac{x}{c}\right)\cdot \frac{1}{c}=\frac{1}{c}{f}_X\left(\frac{x}{c}\right)=\frac{1}{c}\lambda e^{-\lambda \frac{x}{c}}$

$\text{Finally, we have obtained that for}x\ge 0\text{we have that}$

$f_{c}X\left(x\right)=\frac{\lambda }{c}{e}^{-\frac{\lambda }{c}x}$

$\text{and otherwise it is equal to zero. Hence, we have proved that}cX~\mathrm{Expo}\left(\frac{\lambda }{c}\right)$

Final Answer

With differentiating, we have that,

$\frac{d}{dx}{F}_{cX}\left(x\right)=\frac{d}{dx}{F}_X\left(\frac{x}{c}\right)\cdot \frac{1}{c}=\frac{1}{c}{f}_X\left(\frac{x}{c}\right)=\frac{1}{c}\lambda e^{-\lambda \frac{x}{c}}$

$\text{Finally, we have obtained that for}x\ge 0\text{we have that}$

$f_{c}X\left(x\right)=\frac{\lambda }{c}{e}^{-\frac{\lambda }{c}x}$