Distribution of minimum of random variable and its square

Question

Suppose $X \sim U[0,2]$ is a uniformly distributed random variable. What are the distributions of $\min\{X,X^2\}$ and $\max\{1,X\}?$

The density function of $X$ has to be $f(x)=\frac{1}{2}$ and $$F_{X^2}(x)=P_{X^2}(X^2\leq x)=P_{X^2}(-\sqrt{x}\leq X \leq\sqrt{x})=P_{X^2}(X\leq x)-P_{X^2}(X<-\sqrt{x})=F_{X^2}(\sqrt{x})-F_{X^2}(-\sqrt{x}).$$

No idea how to go further.

Answer 1

$$F_{\max\{1,X\}}(x)=P(\max\{1,X\}\le x) \tag{1}$$

$$=P(1 \le x \cap X \le x) \tag{2}$$

$$=P(1 \le x) \times P(X \le x) \tag{3}$$

$$=1_{1 \le x}(x) \times P(X \le x) \tag{4}$$

$$=1_{1 \le x}(x) \times F_X(x) \tag{5}$$

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Remarks and explanations:

$(1)$ Say $P_{X^2}(x)$ or $P(X_2 \le x)$ but not $P_{X^2}(X^2 \le x)$

$(2)$ The elder of two people is younger than a third person iff the third person is older than the two people.

$(3)$ 1 and X are independent for any X! This is because constant or a.s. constant random variables are independent of any random variable, including themselves. This because events of probability 0 or 1 are independent of any other event, including themselves.

$(4)$ x isn't random. It is predetermined. So it either is or isn't greater than 1. Note that this and the next indicator functions are deterministic, i.e. functions of $x$ and not of $\omega$

$(5)$ If $x < 1$, then $\max\{1,X\} > x$: I'm almost sure of it. If $x \ge 1$, then making $X$ become $1$ if $X$ happens to fall below 1 doesn't make $X$ more or less likely to be $\le x$.

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$$F_{\min\{X^2,X\}}(x)=P(\min\{X^2,X\}\le x)$$

$$=P(X^2\le x \cup X\le x)$$

$$=P(X^2\le x) + P(X\le x) - P(X^2\le x \cap X\le x)$$

Now $P(X^2\le x \cap X\le x)$

$$ = P(X^2\le x \cap X\le x \cap X^2 \le X) + P(X^2\le x \cap X\le x \cap X^2 > X)$$

$$ = P(X\le x \cap X^2 \le X) + P(X^2\le x \cap X^2 > X)$$

$$ = P(X\le x \cap 0 < X < 1) + P(X^2\le x \cap 1 < X < 2)$$

$$ = P(X \in (-\infty,x) \cap (0,1)) + P(X^2\le x \cap 1 < X < 2)$$

$$ = P(X \in (-\infty,\min\{x,1\})) + P(X\le \sqrt{x} \cap 1 < X < 2)$$

$$ = F_X(\min\{x,1\}) + P(X\le \sqrt{x} \cap 1 < X < 2)$$

Now $P(X \le \sqrt{x} \cap 1 < X < 2)$

For $x \le 1$ or $x \ge 4$, $$P(X\le \sqrt{x} \cap 1 < X < 2) = P(X \in \emptyset^{\mathbb R}) = P(\emptyset^{\Omega}) = 0$$

For $1 < x < 2$, $$P(X\le \sqrt{x} \cap 1 < X < 2) = P(X\le \sqrt{x})$$

Answer 2

Let $Y = \min\{X,X^2\}$. Note that $Y = X^2$ when $0 \le X \le 1$, and $Y = X$ when $1 \le X \le 2$. Also, $\Pr\{X \le x\} = x/2$ for $0 \le x \le 2$. Using this, we get $$\Pr\{Y \le y\} = \Pr\{X^2 \le y\} = \Pr\{X \le \sqrt{y}\} = \sqrt{y}/2$$ for $0 \le y \le 1$. And $$\Pr\{Y \le y\} = \Pr\{X \le y\} = y/2$$ for $1 \le y \le 2$.

Use the same idea to compute the CDF of $\max\{1,X\}$. Once you have the CDFs, finding the PDFs should be straightforward.