Find the cumulative distribution function of $Y=min(1,X)$ for $f(x)= \frac 1 2x$

Question

Let $X$ be a random variable with pdf $$f(x)= \frac 1 2x$$ with $x \in [0,2]$.

Find the cumulative distribution function of $Y=min(1,X)$.

Attempt: $\mathbb P(min(1,X)\leq t) = \mathbb P(X\geq1)\mathbb P(t\geq 1) + \mathbb P(X\geq1)\mathbb P(t\geq 1) = 0 + F(1)F(t) = \frac{t^2}{16}$ which is not a CDF.

Answer 1

If $0\leq t \leq 1$ we have $\mathbb P(min(1,X)\leq t) = \mathbb P(X\leq t) = F(t)$.

If $1\leq t \leq 2$ then $\mathbb P(min(1,X)\leq t) = 1$.

Note that $F(1) = \frac 1 4$ so that the CDF is not continuos and something wierd is going on... Probably $\mathbb P(Y=1) = 1 - F(1) = \frac 3 4$.

Answer 2

$$ \def\P{\mathbb{P}} $$

If $t < 0$, we have $$ \mathbb{P}(Y \leq t) = 0. $$ If $0 < t < 1$, we have \begin{align*} \mathbb{P}(Y \leq t) &= \mathbb{P}(\min(1, X) \leq t)\\ &= \mathbb{P}(X \leq t) \\ &= \frac{t^2}{4}. \end{align*}

If $t \geq 1$, we have \begin{align*} \mathbb{P}(Y \leq t) &= \mathbb{P}(\min(1, X) \leq t) \\ &= 1.\\ \end{align*}

The discontinuity of the CDF reflects the positive probability, which is $3/4$, that $Y = 1$.