Exponential CDF to a higher power of $X$

Question

$X$ follwing an exponential distribution

I am requested to compute $P(X^3\le x)$

What I do is to take the pdf, substitute like this $P(( Y=X^3) \le x)$

And end up with this pdf= $\int \frac{1}{3x^{2}}\lambda e^{-\lambda x^{3}}$

Is this correct?

Answer 1

Guide:

Taking cube root,

\begin{align} P(X^3 \le x) &= P(X \le x^\frac13) \end{align}

Now you just have to use the CDF of $X$ to compute this quantity.

Answer 2

You seem to have attempted to do this:$$\begin{align} f_{X^n}(x) &=\lvert\dfrac{\mathrm d ~~}{\mathrm d x}\mathsf P(X^n\leq x)\rvert \\ &=\lvert\dfrac{\mathrm d ~~}{\mathrm d x}\mathsf P(X\leq x^{1/n})\rvert &&\text{if $n\in\Bbb N^+$} \\ &=\lvert\dfrac{\mathrm d ~~}{\mathrm d x}\int_0^{x^{1/n}} \lambda \exp(-\lambda~s)~\mathrm d s\rvert\cdot\mathbf 1_{x\in(0;\infty)} \end{align}$$ Next, apply the universal principle of calculus.

$$\begin{align}f_{X^n}(x) &=\lvert\dfrac{\mathrm d x^{1/n}}{\mathrm d x}\rvert\cdot\lambda \exp(-\lambda~x^{1/n})\cdot\mathbf 1_{x\in(0;\infty)}\end{align}$$