Calculating pdf of minimum of i.i.d. random variables with threshold condition Let say I have $m$ i.i.d uniform random variables $U_1, U_2,...U_m$ that range between 0 and 1. I generate $m$ number by using each of random variable and select the one which is minimum among the numbers that exceed threshold. My goal is to obtain pdf of minimum of random variables that exceed a threshold $\gamma$.
Let me rephrase,

*

*I generate $m$ number by using uniform distribution $\sim U\left(0,1\right)$.

*I select the numbers which are bigger than $\gamma$.

*I select the minimum number.

How can I find the pdf of this number? In my opinion, the problem is not easy as it seems.

Let $\ G=\big|\{\,i\ |\,U_i<\gamma\,\}\big|\ $ and $\ V\ $ be the value of the number you choose in step $3$. Then $\ \mathbb{P}\big(G=g\big)=$$\,{m\choose g}\gamma^g(1-\gamma)^{m-g}\ $, and given that $\ G=g\ $ there are $\ m-g\ $ of the variates $\ U_1,U_2,\dots,U_m\ $ that will be uniformly distributed over the interval $\ [\gamma,1]\ $, and the minimum of them will be greater than $\ x\in[\gamma,1]\ $ if and only if all $\ m-g\ $ of them are.  Therefore $\ \mathbb{P}\big(V> x\,|\,G=g\big)=$$\,\left(\frac{1-x}{1-\gamma}\right)^{m-g}\ $ for $\ x\in[\gamma,1]\ $ and $\ g=0,1,\dots, m-1\ $. Therefore
\begin{align}
\mathbb{P}\big(V> x\big)&=\sum_{g=0}^{m-1}\mathbb{P}\big(V> x\,|\,G=g\big)\mathbb{P}\big(G=g\big)\\
&=\sum_{g=0}^{m-1}{m\choose g}\gamma^g(1-x)^{m-g}\\
&=(1+\gamma-x)^m-\gamma^m\ .
\end{align}
Hence
$$
\mathbb{P}\big(\{V\le x\}\cup\{V\ \text{is undefined}\}\big)=1+\gamma^m-(1+\gamma-x)^m\ .
$$
If we take the probability, $\ \gamma^m\ $, of $\ V$'s being undefined as negligible, as the OP indicates as being the case in a comment, we get
$$
\mathbb{P}\big(V\le x\big)=1-(1+\gamma-x)^m
$$
for the cumulative distribution function of $\ V\ $, and
$$
p_V(x)=m(1+\gamma-x)^{m-1}
$$
for its density function.
