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Let $(U_n)_n$ be a sequence of i.i.d random variables such that $U_{n} \sim U[0,1] \forall n$. Define $Y_{n}=\max\{U_{1}, \frac{U_{2}}{2}, \ldots, \frac{U_{n}}{n}\}$, prove that $Y_n$ converges in distribution to a variable $Y$, whose law must identify.

Hint: Investigate beta distribution.

My attempt: My idea was to use the distribution of the maximum in this case, thus obtaining that, for $n\in \mathbb{N}$:

$$\mathbb{P}(Y_{n}\leq y)=\mathbb{P}\left(\bigcap_{i=1}^{n}\{\frac{U_{i}}{i}\leq y\}\right)=\prod_{i=1}^{n}\mathbb{P}\left(\frac{U_i}{i}\leq y\right)$$

It follows: $$\mathbb{P}(Y_{1}\leq y)=y1_{[0,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{2}\leq y)=2y^{2}1_{[0,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{3}\leq y)=6y^{3}1_{[0,1/3]}(y)+2y^{2}1_{(1/3,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\vdots$$ Assuming that this development is correct. What would be the expression for $\mathbb{P}(Y_{n}\leq y)$? Does $Y_n$ converge in distribution to a beta random variable?

Actualization: I obtain $\forall n\geq 2$: $$\mathbb{P}(Y_{n}\leq y)=n!y^{n}1_{[0,1/n]}(y)+\sum_{k=1}^{n}k!y^{k}1_{\left(\frac{1}{k+1},\frac{1}{k}\right]}(y) + 1_{(1,+\infty)}(y)$$

But I can't see to which distribution this expression should converge.

Let $(U_n)_n$ be a sequence of i.i.d random variables such that $U_{n} \sim U[0,1] \forall n$. Define $Y_{n}=\max\{U_{1}, \frac{U_{2}}{2}, \ldots, \frac{U_{n}}{n}\}$, prove that $Y_n$ converges in distribution to a variable $Y$, whose law must identify.

Hint: Investigate beta distribution.

My attempt: My idea was to use the distribution of the maximum in this case, thus obtaining that, for $n\in \mathbb{N}$:

$$\mathbb{P}(Y_{n}\leq y)=\mathbb{P}\left(\bigcap_{i=1}^{n}\{\frac{U_{i}}{i}\leq y\}\right)=\prod_{i=1}^{n}\mathbb{P}\left(\frac{U_i}{i}\leq y\right)$$

It follows: $$\mathbb{P}(Y_{1}\leq y)=y1_{[0,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{2}\leq y)=2y^{2}1_{[0,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{3}\leq y)=6y^{3}1_{[0,1/3]}(y)+2y^{2}1_{(1/3,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\vdots$$ Assuming that this development is correct. What would be the expression for $\mathbb{P}(Y_{n}\leq y)$? Does $Y_n$ converge in distribution to a beta random variable?

Actualization: I obtain $\forall n\geq 2$: $$\mathbb{P}(Y_{n}\leq y)=n!y^{n}1_{[0,1/n]}(y)+\sum_{k=1}^{n-1}k!y^{k}1_{\left(\frac{1}{k+1},\frac{1}{k}\right]}(y) + 1_{(1,+\infty)}(y)$$

But I can't see to which distribution this expression should converge.

Let $(U_n)_n$ be a sequence of i.i.d random variables such that $U_{n} \sim U[0,1] \forall n$. Define $Y_{n}=\max\{U_{1}, \frac{U_{2}}{2}, \ldots, \frac{U_{n}}{n}\}$, prove that $Y_n$ converges in distribution to a variable $Y$, whose law must identify.

Hint: Investigate beta distribution.

My attempt: My idea was to use the distribution of the maximum in this case, thus obtaining that, for $n\in \mathbb{N}$:

$$\mathbb{P}(Y_{n}\leq y)=\mathbb{P}\left(\bigcap_{i=1}^{n}\{\frac{U_{i}}{i}\leq y\}\right)=\prod_{i=1}^{n}\mathbb{P}\left(\frac{U_i}{i}\leq y\right)$$

It follows: $$\mathbb{P}(Y_{1}\leq y)=y1_{[0,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{2}\leq y)=2y^{2}1_{[0,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{3}\leq y)=6y^{3}1_{[0,1/3]}(y)+2y^{2}1_{(1/3,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\vdots$$ Assuming that this development is correct. What would be the expression for $\mathbb{P}(Y_{n}\leq y)$? Does $Y_n$ converge in distribution to a beta random variable?

Let $(U_n)_n$ be a sequence of i.i.d random variables such that $U_{n} \sim U[0,1] \forall n$. Define $Y_{n}=\max\{U_{1}, \frac{U_{2}}{2}, \ldots, \frac{U_{n}}{n}\}$, prove that $Y_n$ converges in distribution to a variable $Y$, whose law must identify.

Hint: Investigate beta distribution.

My attempt: My idea was to use the distribution of the maximum in this case, thus obtaining that, for $n\in \mathbb{N}$:

$$\mathbb{P}(Y_{n}\leq y)=\mathbb{P}\left(\bigcap_{i=1}^{n}\{\frac{U_{i}}{i}\leq y\}\right)=\prod_{i=1}^{n}\mathbb{P}\left(\frac{U_i}{i}\leq y\right)$$

It follows: $$\mathbb{P}(Y_{1}\leq y)=y1_{[0,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{2}\leq y)=2y^{2}1_{[0,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{3}\leq y)=6y^{3}1_{[0,1/3]}(y)+2y^{2}1_{(1/3,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\vdots$$ Assuming that this development is correct. What would be the expression for $\mathbb{P}(Y_{n}\leq y)$? Does $Y_n$ converge in distribution to a beta random variable?

Actualization: I obtain $\forall n\geq 2$: $$\mathbb{P}(Y_{n}\leq y)=n!y^{n}1_{[0,1/n]}(y)+\sum_{k=1}^{n}k!y^{k}1_{\left(\frac{1}{k+1},\frac{1}{k}\right]}(y) + 1_{(1,+\infty)}(y)$$

But I can't see to which distribution this expression should converge.

Let $(U_n)_n$ be a sequence of i.i.d random variables such that $U_{n} \sim U[0,1] \forall n$. Define $Y_{n}=\max\{U_{1}, \frac{U_{2}}{2}, \ldots, \frac{U_{n}}{n}\}$, prove that $Y_n$ converges in distribution to a variable $Y$, whose law must identify.

Hint: Investigate beta distribution.

My attempt: My idea was to use the distribution of the maximum in this case, thus obtaining that, for $n\in \mathbb{N}$ and $y\in [0,1/n]$:

$$\mathbb{P}(Y_{n}\leq y)=\mathbb{P}\left(\bigcap_{i=1}^{n}\{\frac{U_{i}}{i}\leq y\}\right)=\prod_{i=1}^{n}\mathbb{P}\left(\frac{U_i}{i}\leq y\right)=n!y^{n}1_{[0,1/n)}+1_{[1/n,\infty)}$$

Assuming that this development is correct. How could convergence be verified in distribution to a beta random variable? Is it really true that $Y_n$ converges in distribution to a beta random variable?

Let $(U_n)_n$ be a sequence of i.i.d random variables such that $U_{n} \sim U[0,1] \forall n$. Define $Y_{n}=\max\{U_{1}, \frac{U_{2}}{2}, \ldots, \frac{U_{n}}{n}\}$, prove that $Y_n$ converges in distribution to a variable $Y$, whose law must identify.

Hint: Investigate beta distribution.

My attempt: My idea was to use the distribution of the maximum in this case, thus obtaining that, for $n\in \mathbb{N}$:

$$\mathbb{P}(Y_{n}\leq y)=\mathbb{P}\left(\bigcap_{i=1}^{n}\{\frac{U_{i}}{i}\leq y\}\right)=\prod_{i=1}^{n}\mathbb{P}\left(\frac{U_i}{i}\leq y\right)$$

It follows: $$\mathbb{P}(Y_{1}\leq y)=y1_{[0,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{2}\leq y)=2y^{2}1_{[0,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\mathbb{P}(Y_{3}\leq y)=6y^{3}1_{[0,1/3]}(y)+2y^{2}1_{(1/3,1/2]}(y)+y1_{(1/2,1]}(y)+1_{(1,+\infty)}(y)$$ $$\vdots$$ Assuming that this development is correct. What would be the expression for $\mathbb{P}(Y_{n}\leq y)$? Does $Y_n$ converge in distribution to a beta random variable?