# Finding CDF of $Z$ with PDF of $X$

Suppose the random variable $$X$$ has PDF
$$f_X(x) = 2x^2~, \quad\text{when}~ 0 < x < 1~,$$ and zero otherwise.
Suppose $$Z = 4X+9$$, what is the CDF of $$Z$$? Then find the PDF of $$Z$$.

I began by finding $$X=(Z-9)/4$$, and then finding the integral of $$f_X(x)$$. I'm not sure of next steps.

• What trouble are you having with this? Please show your thoughts on , and attempts at , finding a solution. Apr 26 '19 at 2:56
• It looks like $f_x(x)$ should be $3x^2$, since $\int_0^1f_x(x)dx$ must $=1$. Apr 26 '19 at 3:34
• @herbsteinberg Indeed. Good catch. Apr 26 '19 at 3:35

I began by finding X=(Z-9)/4, and then finding the integral of fx(x). I'm not sure of next steps.

That will give you the CDF of $$Z$$.

To find the pdf, differentiate with respect to $$z$$.\begin{align}{F}_Z(z)&={F}_X\left[(z-9)/4\right]\\[1ex]&=\int_{-\infty}^{(z-9)/4} f_X(x)~\mathrm d x\\[1ex]&=\mathbf 1_{0\leq (z-9)\le 4}~\int_0^{(z-9)/4} \require{cancel}\cancelto3{\color{red}2}x^2~\mathrm d x+\mathbf 1_{4\leq (z-9)}\\[1ex]&=\lower{2ex}\ddots\\[3ex] {f}_Z(x)&=\dfrac{\mathrm d~~}{\mathrm d z}F_Z(z)\\[1ex]&=\lower{2ex}\ddots\end{align}

• This result did not take into account the error in the original statement. Apr 26 '19 at 3:50
• I see the correction. Apr 26 '19 at 16:02

$$P(Z\le z)=P(4X+9\le z)=P(X\le \frac{z-9}{4})=(\frac{z-9}{4})^3$$ (the CDF of Z) for $$9\le z \le 13$$ The CDF $$=0$$ for smaller $$z$$ and $$=1$$ for larger $$z$$.

The PDF of $$z$$ is $$\frac{3}{4}(\frac{z-9}{4})^2$$ for $$9\le z \le 13$$ and $$=0$$ otherwise.