# Functions of continuous random variables

Let Y be an exponential random variable with parameter $\tau > 0$. Compute the cdf and pdf of $F_W$ where $W = Y^3$

The solution states the cdf as $1 - e^{\frac{-y^\frac{1}{3}}{t}}$ because $F_Y =1 - e^{\frac{y}{\tau}}$. $P[W \leq w] = P[Y^3 \leq w] = P[Y \leq y^{\frac{1}{3}}] = 1 - e^{\frac{-y^\frac{1}{3}}{t}}$

I tried achieving the same results with the following logic: $P[W \leq w] = P[Y^3 \leq w] = P[Y \leq w^{\frac{1}{3}}] = \int_{0}^{w^\frac{1}{3}} \frac{1}{\tau} e^{-\frac{y}{\tau}}dy =\left(e^{-\frac{\sqrt[3]{w}}{\tau}}-1\right)\tau+\frac{\sqrt[3]{w}}{\tau}$

However this does not return the same result. Does anyone know why my method did not return the correct result?

• I don't understand what you're doing during your integration. $\int_0^a be^{-bw}dw=1-e^{-ba}$. Where are you getting $x$ and where are you getting the second term? Oct 10, 2017 at 22:37
• @AlexR. I've edited my logic a bit. Let me know if there is more confusion Oct 10, 2017 at 22:42
• Again, you are evaluating the integral incorrectly. Oct 10, 2017 at 22:52
• That means you entered it incorrectly into symbolab. According to you if $w$ goes to infinity, then probability goes to infinity as well. Does that sound right to you? Oct 10, 2017 at 22:57
• You should not rely on automatic computation you can't apply reasonableness checks to; otherwise your usage errors turn into gospel. ( (along with the much less likely but still possible presence of bugs). Oct 11, 2017 at 0:03

First of all,

$$P[W \le w] = P[Y^3 \le w] = P[Y \le w^{1/3}] = 1- e^\frac{-w^{1/3} }{t}$$

ie. this is a function of W not Y.

$$\int_0^{w^{1/3}} \frac{1}{t} e^{-y/t} dy$$
Let $$u = \frac{-y}{t}$$
$$- \int_0^{-\frac{w^{1/3}}{t}} e^u du$$