Distribution of $min(x,\frac{1}{2})$

Question

We have a random variable $X$ which is uniformly $U(0,1)$ distributed. It is required to obtain the PDF of $min(x,\frac{1}{2})$.

My approach

If $x<\frac{1}{2}$, Y = x If $x\geq\frac{1}{2}$, Y = \frac{1}{2}

I am not able to translate the above information into the PDF. Any help would be appreciated. Thanks in advance.

Answer 1

Let $Y=X\wedge\frac12$, then $$Y = X\cdot\mathsf 1_{\left(0,\frac12\right)}(X) + \frac12\cdot \mathsf 1_{\left[\frac12,1\right)}(X).$$ It follows that \begin{align} \mathbb P(Y\leqslant y) &= \mathbb P\left(Y\leqslant y, 0<X<\frac12 \right) + \mathbb P\left(Y\leqslant y, \frac12\leqslant X<1 \right)\\ &= y\cdot \mathsf 1_{\left(0,\frac12\right)}(y) + \mathsf 1_{\left[\frac12,\infty\right)}(y). \end{align}