Convergence in distribution of a product












1














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







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  • 1




    You probably mean $V_nto 0$ in probability.
    – Davide Giraudo
    Nov 29 '18 at 10:03
















1














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







share|cite|improve this question




















  • 1




    You probably mean $V_nto 0$ in probability.
    – Davide Giraudo
    Nov 29 '18 at 10:03














1












1








1


1





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 10:03









Davide Giraudo

125k16150260




125k16150260










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














  • 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










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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.






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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.






    share|cite|improve this answer


























      1














      $P{V_n >t}=e^{-nt}$ 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.






      share|cite|improve this answer
























        1












        1








        1






        $P{V_n >t}=e^{-nt}$ 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.






        share|cite|improve this answer












        $P{V_n >t}=e^{-nt}$ 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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 5:27









        Kavi Rama Murthy

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