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Finding the distribution of $W$ explicitly

Limit of $(1+ x/n)^n$ when $n$ tends to infinityFinding distribution $W$Very Important question. Limiting distributionSufficient Statistics and Maximum LikelihoodFind the limiting distribution of the following random variableFnd a sequence to be convergence in distributionFinding the limiting distribution $nmin(X_1, dots , X_n)$ with uniformly distributed $X_i$Proof that this distribution is N(0,1)?Sufficient statistic for Uniform distribution.Finding distribution of $fracX_1+X_2 X_3sqrt1+X_3^2$Conditional distribution of life time for a renewal process

0

$begingroup$

Let $X_1,ldots, X_n sim U[0, 1]$. Let $Y_n = min_1 leq ileq n X_i$.
Show that $nY_n$ converges in distribution to some random variable $W$ . Find
the distribution of $W$ explicitly.

I know that $$F(W)=P(Wle w)=P(nYnle w)=P(Ynle w/n)=1-(1-w/n)^n$$

But I cannot continue to finish the question.

statistics probability-distributions

share|cite|improve this question

edited Mar 20 at 19:36

StubbornAtom

6,29831440

asked Mar 20 at 2:05

aaaaaaaa

11

$endgroup$

add a comment | 

0

$begingroup$

Let $X_1,ldots, X_n sim U[0, 1]$. Let $Y_n = min_1 leq ileq n X_i$.
Show that $nY_n$ converges in distribution to some random variable $W$ . Find
the distribution of $W$ explicitly.

I know that $$F(W)=P(Wle w)=P(nYnle w)=P(Ynle w/n)=1-(1-w/n)^n$$

But I cannot continue to finish the question.

statistics probability-distributions

share|cite|improve this question

edited Mar 20 at 19:36

StubbornAtom

6,29831440

asked Mar 20 at 2:05

aaaaaaaa

11

$endgroup$

add a comment | 

$begingroup$

Let $X_1,ldots, X_n sim U[0, 1]$. Let $Y_n = min_1 leq ileq n X_i$.
Show that $nY_n$ converges in distribution to some random variable $W$ . Find
the distribution of $W$ explicitly.

I know that $$F(W)=P(Wle w)=P(nYnle w)=P(Ynle w/n)=1-(1-w/n)^n$$

But I cannot continue to finish the question.

statistics probability-distributions

share|cite|improve this question

edited Mar 20 at 19:36

StubbornAtom

6,29831440

asked Mar 20 at 2:05

aaaaaaaa

11

$endgroup$

share|cite|improve this question

edited Mar 20 at 19:36

StubbornAtom

6,29831440

asked Mar 20 at 2:05

aaaaaaaa

11

share|cite|improve this question

edited Mar 20 at 19:36

StubbornAtom

6,29831440

asked Mar 20 at 2:05

aaaaaaaa

11

edited Mar 20 at 19:36

StubbornAtom

6,29831440

edited Mar 20 at 19:36

StubbornAtom

6,29831440

asked Mar 20 at 2:05

aaaaaaaa

11

asked Mar 20 at 2:05

aaaaaaaa

11

2 Answers
2

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0

$begingroup$

Hint: Use the fact that $limlimits_ntoinftyleft(1 +fracxnright)^n = e^x$ for any $xinmathbbR$.

share|cite|improve this answer

answered Mar 20 at 2:13

Minus One-TwelfthMinus One-Twelfth

2,933413

$endgroup$

add a comment | 

0

$begingroup$

Slightly less elementary than the other answer, but I believe that these facts are also useful to know. I'll give the outline of the proof, necessary steps aren't that hard to do:

1) If $ X_i sim U(0, 1)$ and iid, then $min X_i sim B(1, n)$, which is Beta random variable.

2) It can be shown using convergence of moments that $n B(1,n) rightarrow G(1, 1)$, where $G(alpha, beta)$ is Gamma random variable.

share|cite|improve this answer

answered Mar 21 at 6:35

defenestratordefenestrator

334

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add a comment | 

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0

$begingroup$

Hint: Use the fact that $limlimits_ntoinftyleft(1 +fracxnright)^n = e^x$ for any $xinmathbbR$.

share|cite|improve this answer

answered Mar 20 at 2:13

Minus One-TwelfthMinus One-Twelfth

2,933413

$endgroup$

add a comment | 

0

$begingroup$

Hint: Use the fact that $limlimits_ntoinftyleft(1 +fracxnright)^n = e^x$ for any $xinmathbbR$.

share|cite|improve this answer

answered Mar 20 at 2:13

Minus One-TwelfthMinus One-Twelfth

2,933413

$endgroup$

add a comment | 

$begingroup$

Hint: Use the fact that $limlimits_ntoinftyleft(1 +fracxnright)^n = e^x$ for any $xinmathbbR$.

share|cite|improve this answer

answered Mar 20 at 2:13

Minus One-TwelfthMinus One-Twelfth

2,933413

$endgroup$

share|cite|improve this answer

answered Mar 20 at 2:13

Minus One-TwelfthMinus One-Twelfth

2,933413

answered Mar 20 at 2:13

Minus One-TwelfthMinus One-Twelfth

2,933413

answered Mar 20 at 2:13

Minus One-TwelfthMinus One-Twelfth

2,933413

0

$begingroup$

Slightly less elementary than the other answer, but I believe that these facts are also useful to know. I'll give the outline of the proof, necessary steps aren't that hard to do:

1) If $ X_i sim U(0, 1)$ and iid, then $min X_i sim B(1, n)$, which is Beta random variable.

2) It can be shown using convergence of moments that $n B(1,n) rightarrow G(1, 1)$, where $G(alpha, beta)$ is Gamma random variable.

share|cite|improve this answer

answered Mar 21 at 6:35

defenestratordefenestrator

334

$endgroup$

add a comment | 

0

$begingroup$

Slightly less elementary than the other answer, but I believe that these facts are also useful to know. I'll give the outline of the proof, necessary steps aren't that hard to do:

1) If $ X_i sim U(0, 1)$ and iid, then $min X_i sim B(1, n)$, which is Beta random variable.

2) It can be shown using convergence of moments that $n B(1,n) rightarrow G(1, 1)$, where $G(alpha, beta)$ is Gamma random variable.

share|cite|improve this answer

answered Mar 21 at 6:35

defenestratordefenestrator

334

$endgroup$

add a comment | 

$begingroup$

Slightly less elementary than the other answer, but I believe that these facts are also useful to know. I'll give the outline of the proof, necessary steps aren't that hard to do:

1) If $ X_i sim U(0, 1)$ and iid, then $min X_i sim B(1, n)$, which is Beta random variable.

2) It can be shown using convergence of moments that $n B(1,n) rightarrow G(1, 1)$, where $G(alpha, beta)$ is Gamma random variable.

share|cite|improve this answer

answered Mar 21 at 6:35

defenestratordefenestrator

334

$endgroup$

share|cite|improve this answer

answered Mar 21 at 6:35

defenestratordefenestrator

334

answered Mar 21 at 6:35

defenestratordefenestrator

334

answered Mar 21 at 6:35

defenestratordefenestrator

334