If $X_1,..,X_n$ are independent Uniform$(0,\theta)$ variables, show that $Y_n=\text{min}\Big(X_1,..,X_n\Big)$ has density $f_{Y_n}(y)=\frac{n}{\theta}\Big(1-\frac{y}{\theta}\Big)^{n-1} \ \ \ y\in(0,\theta)$.
Attempt:
Computing the CDF:
\begin{align}
F_{Y_n}(y)&=\mathbb{P}(Y_n\leq y) \\
&= \mathbb{P}(\text{min}(X_1,..,X_n)\leq y) \\
&= 1-\mathbb{P}(X_1>y,X_2>y,..,X_n>y) \\
&= 1-\mathbb{P}(X_1>y)\mathbb{P}(X_2>y)…\mathbb{P}(X_n>y) \ \ \ \ \ \text{(by independence)}\\
\end{align}
Next I tried to calculate $\mathbb{P}(X_i>y) \ \ \ (i=1,..,n)$. Hence,
$$\int_{y}^{\theta} f_X(t) \ dt=\int_{y}^{\theta} \frac{1}{t} \ dt=\text{ln}(\theta)-\text{ln}(y)$$
Thus $$F_{Y_n}(y)=1-\Big(\text{ln}(\theta)-\text{ln}(y)\Big)^n$$
Differentiating, $$f_{Y_n}(y)=\frac{n}{y}\Big(\text{ln}(\theta)-\text{ln}(y)\Big)^{n-1}$$
Where have I gone so wrong? As a prelude to this question, I showed using this method that $F_{Y_n}(y)=1-\Big(1-\frac{y}{n}\Big)^n \ \ \ \ y\in(0,\theta)$ where $X_1,..,X_n$ is a sequence of iid Uniform$(0,1)$ random variables and $Y_n=n\text{min}\Big(X_1,..,X_n\Big)$.
Best Answer
Since $X_{i}$ is Uniform$(0,\theta)$, it's probability density function should be
$f_{X_{i}}(x)=\frac{1}{\theta}$ which is a constant. Note your density function varies with values of $t$, which obviously cannot be the case for a uniform distribution.