Convergence in probability: The inverse of the simple mean
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I have a question on convergence: I have to prove that $frac{n}{U_{n}} longrightarrow 1$ in probability, where $U_{n}=sum X_{i}$, $X_{i}sim mathrm{Exp}(1)$ and because of this, $U_{n}sim mathrm{Gamma}(n,1)$.
This problem had two parts, the other part was to prove that $frac{U_{n}}{n}longrightarrow 1$ in probability which I proved invoking the weak law for big numbers. But this, I have no clue how to prove it.
Thanks so much for your help! :)
probability-theory statistics convergence exponential-distribution gamma-distribution
edited Nov 29 '18 at 9:56
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:25
DadadaveDadadave
9118
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add a comment |
$begingroup$
I have a question on convergence: I have to prove that $frac{n}{U_{n}} longrightarrow 1$ in probability, where $U_{n}=sum X_{i}$, $X_{i}sim mathrm{Exp}(1)$ and because of this, $U_{n}sim mathrm{Gamma}(n,1)$.
This problem had two parts, the other part was to prove that $frac{U_{n}}{n}longrightarrow 1$ in probability which I proved invoking the weak law for big numbers. But this, I have no clue how to prove it.
Thanks so much for your help! :)
probability-theory statistics convergence exponential-distribution gamma-distribution
edited Nov 29 '18 at 9:56
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:25
DadadaveDadadave
9118
$endgroup$
add a comment |
$begingroup$
I have a question on convergence: I have to prove that $frac{n}{U_{n}} longrightarrow 1$ in probability, where $U_{n}=sum X_{i}$, $X_{i}sim mathrm{Exp}(1)$ and because of this, $U_{n}sim mathrm{Gamma}(n,1)$.
This problem had two parts, the other part was to prove that $frac{U_{n}}{n}longrightarrow 1$ in probability which I proved invoking the weak law for big numbers. But this, I have no clue how to prove it.
Thanks so much for your help! :)
probability-theory statistics convergence exponential-distribution gamma-distribution
edited Nov 29 '18 at 9:56
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:25
DadadaveDadadave
9118
$endgroup$
I have a question on convergence: I have to prove that $frac{n}{U_{n}} longrightarrow 1$ in probability, where $U_{n}=sum X_{i}$, $X_{i}sim mathrm{Exp}(1)$ and because of this, $U_{n}sim mathrm{Gamma}(n,1)$.
This problem had two parts, the other part was to prove that $frac{U_{n}}{n}longrightarrow 1$ in probability which I proved invoking the weak law for big numbers. But this, I have no clue how to prove it.
Thanks so much for your help! :)
probability-theory statistics convergence exponential-distribution gamma-distribution
probability-theory statistics convergence exponential-distribution gamma-distribution
edited Nov 29 '18 at 9:56
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:25
DadadaveDadadave
9118
edited Nov 29 '18 at 9:56
Davide Giraudo
126k16150261
asked Nov 28 '18 at 21:25
DadadaveDadadave
9118
edited Nov 29 '18 at 9:56
Davide Giraudo
126k16150261
edited Nov 29 '18 at 9:56
Davide Giraudo
126k16150261
edited Nov 29 '18 at 9:56
Davide Giraudo
126k16150261
126k16150261
asked Nov 28 '18 at 21:25
DadadaveDadadave
9118
asked Nov 28 '18 at 21:25
DadadaveDadadave
9118
asked Nov 28 '18 at 21:25
DadadaveDadadave
9118
9118
add a comment |
add a comment |
1 Answer
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I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 '18 at 21:42
angryavianangryavian
40.7k23380
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So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
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– Dadadave
Nov 28 '18 at 22:11
1
$begingroup$
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
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– Stockfish
Nov 29 '18 at 10:09
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 '18 at 21:42
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
$endgroup$
– Dadadave
Nov 28 '18 at 22:11
1
$begingroup$
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
$endgroup$
– Stockfish
Nov 29 '18 at 10:09
add a comment |
$begingroup$
I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 '18 at 21:42
angryavianangryavian
40.7k23380
$endgroup$
$begingroup$
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
$endgroup$
– Dadadave
Nov 28 '18 at 22:11
1
$begingroup$
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
$endgroup$
– Stockfish
Nov 29 '18 at 10:09
add a comment |
$begingroup$
I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 '18 at 21:42
angryavianangryavian
40.7k23380
$endgroup$
I think you can use the continuous mapping theorem on the other part of the question.
answered Nov 28 '18 at 21:42
angryavianangryavian
40.7k23380
answered Nov 28 '18 at 21:42
angryavianangryavian
40.7k23380
answered Nov 28 '18 at 21:42
angryavianangryavian
40.7k23380
answered Nov 28 '18 at 21:42
angryavianangryavian
40.7k23380
40.7k23380
$begingroup$
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
$endgroup$
– Dadadave
Nov 28 '18 at 22:11
1
$begingroup$
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
$endgroup$
– Stockfish
Nov 29 '18 at 10:09
add a comment |
$begingroup$
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
$endgroup$
– Dadadave
Nov 28 '18 at 22:11
1
$begingroup$
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
$endgroup$
– Stockfish
Nov 29 '18 at 10:09
$begingroup$
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
$endgroup$
– Dadadave
Nov 28 '18 at 22:11
$begingroup$
So I just have to use $g(x)=x^{-1}$, and then, since $frac{U_{n}}{n}$ converges in probability to 1, then $g( frac{U_{n}}{n})= frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :)
$endgroup$
– Dadadave
Nov 28 '18 at 22:11
1
1
$begingroup$
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
$endgroup$
– Stockfish
Nov 29 '18 at 10:09
$begingroup$
Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$)
$endgroup$
– Stockfish
Nov 29 '18 at 10:09
add a comment |
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