Convergence in distribution of a product
I'm kinda lost in this kind of problems, so I apologize if this is to easy.
Let $V_nsimoperatorname{Exp}(n)$. I've already prove that $V_n rightarrow 1$ in probability, $e^{-V_n}rightarrow 1$ in probability but I cant manage to prove that $nV_{n}rightarrow V$ in distribution where $Vsimoperatorname{Exp}(1)$.
Hope you guys could help, thanks so much in advance. :)
probability probability-theory statistics convergence exponential-distribution
edited Nov 29 '18 at 10:03
Davide Giraudo
125k16150260
asked Nov 28 '18 at 4:01
Dadadave
9118
add a comment |
I'm kinda lost in this kind of problems, so I apologize if this is to easy.
Let $V_nsimoperatorname{Exp}(n)$. I've already prove that $V_n rightarrow 1$ in probability, $e^{-V_n}rightarrow 1$ in probability but I cant manage to prove that $nV_{n}rightarrow V$ in distribution where $Vsimoperatorname{Exp}(1)$.
Hope you guys could help, thanks so much in advance. :)
probability probability-theory statistics convergence exponential-distribution
edited Nov 29 '18 at 10:03
Davide Giraudo
125k16150260
asked Nov 28 '18 at 4:01
Dadadave
9118
1
You probably mean $V_nto 0$ in probability.
– Davide Giraudo
Nov 29 '18 at 10:03
add a comment |
I'm kinda lost in this kind of problems, so I apologize if this is to easy.
Let $V_nsimoperatorname{Exp}(n)$. I've already prove that $V_n rightarrow 1$ in probability, $e^{-V_n}rightarrow 1$ in probability but I cant manage to prove that $nV_{n}rightarrow V$ in distribution where $Vsimoperatorname{Exp}(1)$.
Hope you guys could help, thanks so much in advance. :)
probability probability-theory statistics convergence exponential-distribution
edited Nov 29 '18 at 10:03
Davide Giraudo
125k16150260
asked Nov 28 '18 at 4:01
Dadadave
9118
I'm kinda lost in this kind of problems, so I apologize if this is to easy.
Let $V_nsimoperatorname{Exp}(n)$. I've already prove that $V_n rightarrow 1$ in probability, $e^{-V_n}rightarrow 1$ in probability but I cant manage to prove that $nV_{n}rightarrow V$ in distribution where $Vsimoperatorname{Exp}(1)$.
Hope you guys could help, thanks so much in advance. :)
probability probability-theory statistics convergence exponential-distribution
probability probability-theory statistics convergence exponential-distribution
edited Nov 29 '18 at 10:03
Davide Giraudo
125k16150260
asked Nov 28 '18 at 4:01
Dadadave
9118
edited Nov 29 '18 at 10:03
Davide Giraudo
125k16150260
asked Nov 28 '18 at 4:01
Dadadave
9118
edited Nov 29 '18 at 10:03
Davide Giraudo
125k16150260
edited Nov 29 '18 at 10:03
Davide Giraudo
125k16150260
edited Nov 29 '18 at 10:03
Davide Giraudo
125k16150260
125k16150260
asked Nov 28 '18 at 4:01
Dadadave
9118
asked Nov 28 '18 at 4:01
Dadadave
9118
asked Nov 28 '18 at 4:01
Dadadave
9118
9118
1
You probably mean $V_nto 0$ in probability.
– Davide Giraudo
Nov 29 '18 at 10:03
add a comment |
1
You probably mean $V_nto 0$ in probability.
– Davide Giraudo
Nov 29 '18 at 10:03
1
1
You probably mean $V_nto 0$ in probability.
– Davide Giraudo
Nov 29 '18 at 10:03
You probably mean $V_nto 0$ in probability.
– Davide Giraudo
Nov 29 '18 at 10:03
add a comment |
1 Answer
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$P{V_n >t}=e^{-nt}$ by definition of exponential distribution. Hence $P{nV_n >t}=P{V_n >frac t n}=e^{-t}=P{V>t}$. This means $nV_n$ has the same distribution as $V$ for every $n$. Surely, $nV_n to V$ in distribution.
answered Nov 28 '18 at 5:27
Kavi Rama Murthy
51.1k31855
add a comment |
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$P{V_n >t}=e^{-nt}$ by definition of exponential distribution. Hence $P{nV_n >t}=P{V_n >frac t n}=e^{-t}=P{V>t}$. This means $nV_n$ has the same distribution as $V$ for every $n$. Surely, $nV_n to V$ in distribution.
answered Nov 28 '18 at 5:27
Kavi Rama Murthy
51.1k31855
add a comment |
$P{V_n >t}=e^{-nt}$ by definition of exponential distribution. Hence $P{nV_n >t}=P{V_n >frac t n}=e^{-t}=P{V>t}$. This means $nV_n$ has the same distribution as $V$ for every $n$. Surely, $nV_n to V$ in distribution.
answered Nov 28 '18 at 5:27
Kavi Rama Murthy
51.1k31855
add a comment |
$P{V_n >t}=e^{-nt}$ by definition of exponential distribution. Hence $P{nV_n >t}=P{V_n >frac t n}=e^{-t}=P{V>t}$. This means $nV_n$ has the same distribution as $V$ for every $n$. Surely, $nV_n to V$ in distribution.
answered Nov 28 '18 at 5:27
Kavi Rama Murthy
51.1k31855
$P{V_n >t}=e^{-nt}$ by definition of exponential distribution. Hence $P{nV_n >t}=P{V_n >frac t n}=e^{-t}=P{V>t}$. This means $nV_n$ has the same distribution as $V$ for every $n$. Surely, $nV_n to V$ in distribution.
answered Nov 28 '18 at 5:27
Kavi Rama Murthy
51.1k31855
answered Nov 28 '18 at 5:27
Kavi Rama Murthy
51.1k31855
answered Nov 28 '18 at 5:27
Kavi Rama Murthy
51.1k31855
answered Nov 28 '18 at 5:27
Kavi Rama Murthy
51.1k31855
51.1k31855
add a comment |
add a comment |
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You probably mean $V_nto 0$ in probability.
– Davide Giraudo
Nov 29 '18 at 10:03