Let ${X \sim \!\, Exp(\lambda)}$, that is, ${X}$ is exponentially distributed with the probability density function ${f_X(x)= \lambda e^{-\lambda x}}, x≥0$. Determine the density and distribution functions for ${Y:=X^2}$

What i've done so far:

Exponentially distributed function by definition ${F_X(x)= 1- e^{-\lambda x}}$

Let ${g(x)= 1- e^{-\lambda x}}$, which is a continuous monotone strictly growing function.

Now I want to determine the density and distribution functions for ${Y:=g(X^2)=1- e^{-\lambda X}}$

${x^2=y \leftrightarrow x= \sqrt{y}}$ here is where I get stuck. Well, not sure if om on the right track. The answer is ${F_Y(y)= 1- e^{-\lambda \sqrt{y}}}$

And obviously the density function will be the derivative of the distribution function.

What needs to be done here?? thanks on beforehand.


1 Answer 1


Note that if $w\ge 0$ then $$\Pr(X\le w)=\int_0^w \lambda e^{-\lambda x}\,dx=1-e^{-\lambda w}.\tag{1}$$ For $y\ge 0$, we have $$F_Y(y)=\Pr(Y\le y)=\Pr(X^2\le y)=\Pr(X\le \sqrt{y})=1-e^{-\lambda\sqrt{y}}.$$ The last equality is by using (1) with $w=\sqrt{y}$.

Finally the PDF (probability density function) can be obtained by differentiation of this CDF (cumulative density function).

  • $\begingroup$ cheers André!! you delivered last time and you delivered now!!! $\endgroup$ Feb 11, 2015 at 23:46

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