Expectation of Transposed random variable

Question

Suppose that the random variable X is uniformly distributed on the interval [0, 1]) (i.e X ∼ U(0, 1)) distribution and suppose that Z = min$(2, 2X^2 + 1)$ .

(a) Explain why Z does not have a density function.

(b) Find E(Z).

my p.d.f for Z is Z=$2x^2+1$ for $0<x<1$ i don't get why Z doesn't have a pdf and also i dont get how to do part B please help.

Answer 1

A continuous random variable has an individual probability of zero at any one single point.

but according this the probability at x=$1/sqrt(2)$

that makes it discreet at one single point hence the above doesn't have a pdf.

thanks to my friend i found the the solution

Answer 2

I agree that Z does have a density function but it is not what you say, its this:

$$Z=f(x)=\begin{cases} 2x^2+1&,0\le x\le \frac{1}{\sqrt2}\\ 2&,\frac{1}{\sqrt2}\lt x\le1\\ 0&,\text{otherwise} \end{cases}$$

For Part B

$$E(z)=\int_{-\infty}^\infty xf(x)dx$$