Establish bound for a probability using moment generating function












1












$begingroup$


I have the following question



Let




$X_{1}$, $X_{2}$, ..., $X_{n}$ be independent and identically
distributed random variables with moment generating function
$M_{X}(t)$, for -h < t < h, h > 0, $S_{n}$ = $sum_{i=1}^{n}X_i$, and
$bar{X}$ = $S_{n}/n$.




Ultimately I need to establish the following:




Use the fact that $M_{X}(0)=1$ and $dM_{X}(t)/dt$ $vert_{t=0} =
> mathbb{E}(X)$ to show that if $mathbb{E}(X)<0$ then there is a
$0>c<1$ with $P(S_{n}>a) leq c^{n}$. Establish a similar bound for
$P(S_{n} leq a)$.




I have previously shown that




$P(S_{n}>a) leq e^{-at}[M_{X}(t)]^{n}$ for $0<t<h$ , $ageq0$ and $P(S_{n}leq a) leq e^{-at}[M_{X}(t)]^{n}$ for $ -h<t<0 $




As follows:




$P(S_{n} > a) = int_a^{infty} dF(s) leq int_a^{infty} e^{-at} e^{st} dF(s) leq int_{-infty}^{infty} e^{-at} e^{st} dF(s) = e^{-at} mathbb{E} (e^{t sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $




And similarly:




$P(S_{n} leq a) = int_{-infty}^{a} dF(s) leq int_{-infty}^{a} e^{-at} e^{st} dF(s) leq int_{-infty}^{infty} e^{-at} e^{st} dF(s) = e^{-at} mathbb{E} (e^{t sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $




Now, I have really no clue on how to establish a bound at $c_{n}$ using the fact that $M_{X}(t) = 0 $ and $dM_{X}(t)/dt$ $vert_{t=0} = mathbb{E}(X)$ to show that if $mathbb{E}(X)<0$ then there is a $0>c<1$ with $P(S_{n}>a) leq c^{n}$. I have no attempt to show but any ideas would be very welcome.



Thanks in advance






statistics random-variables moment-generating-functions







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I have the following question



    Let




    $X_{1}$, $X_{2}$, ..., $X_{n}$ be independent and identically
    distributed random variables with moment generating function
    $M_{X}(t)$, for -h < t < h, h > 0, $S_{n}$ = $sum_{i=1}^{n}X_i$, and
    $bar{X}$ = $S_{n}/n$.




    Ultimately I need to establish the following:




    Use the fact that $M_{X}(0)=1$ and $dM_{X}(t)/dt$ $vert_{t=0} =
    > mathbb{E}(X)$     to show that if $mathbb{E}(X)<0$ then there is a
    $0>c<1$ with $P(S_{n}>a) leq c^{n}$. Establish a similar bound for
    $P(S_{n} leq a)$.




    I have previously shown that




    $P(S_{n}>a) leq e^{-at}[M_{X}(t)]^{n}$ for $0<t<h$ , $ageq0$ and $P(S_{n}leq a) leq e^{-at}[M_{X}(t)]^{n}$ for $ -h<t<0 $




    As follows:




    $P(S_{n} > a) = int_a^{infty} dF(s) leq int_a^{infty} e^{-at} e^{st} dF(s) leq int_{-infty}^{infty} e^{-at} e^{st} dF(s) = e^{-at} mathbb{E} (e^{t sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $




    And similarly:




    $P(S_{n} leq a) = int_{-infty}^{a} dF(s) leq int_{-infty}^{a} e^{-at} e^{st} dF(s) leq int_{-infty}^{infty} e^{-at} e^{st} dF(s) = e^{-at} mathbb{E} (e^{t sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $




    Now, I have really no clue on how to establish a bound at $c_{n}$ using the fact that $M_{X}(t) = 0 $ and $dM_{X}(t)/dt$ $vert_{t=0} = mathbb{E}(X)$ to show that if $mathbb{E}(X)<0$ then there is a $0>c<1$ with $P(S_{n}>a) leq c^{n}$. I have no attempt to show but any ideas would be very welcome.



    Thanks in advance






    statistics random-variables moment-generating-functions







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I have the following question



      Let




      $X_{1}$, $X_{2}$, ..., $X_{n}$ be independent and identically
      distributed random variables with moment generating function
      $M_{X}(t)$, for -h < t < h, h > 0, $S_{n}$ = $sum_{i=1}^{n}X_i$, and
      $bar{X}$ = $S_{n}/n$.




      Ultimately I need to establish the following:




      Use the fact that $M_{X}(0)=1$ and $dM_{X}(t)/dt$ $vert_{t=0} =
      > mathbb{E}(X)$       to show that if $mathbb{E}(X)<0$ then there is a
      $0>c<1$ with $P(S_{n}>a) leq c^{n}$. Establish a similar bound for
      $P(S_{n} leq a)$.




      I have previously shown that




      $P(S_{n}>a) leq e^{-at}[M_{X}(t)]^{n}$ for $0<t<h$ , $ageq0$ and $P(S_{n}leq a) leq e^{-at}[M_{X}(t)]^{n}$ for $ -h<t<0 $




      As follows:




      $P(S_{n} > a) = int_a^{infty} dF(s) leq int_a^{infty} e^{-at} e^{st} dF(s) leq int_{-infty}^{infty} e^{-at} e^{st} dF(s) = e^{-at} mathbb{E} (e^{t sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $




      And similarly:




      $P(S_{n} leq a) = int_{-infty}^{a} dF(s) leq int_{-infty}^{a} e^{-at} e^{st} dF(s) leq int_{-infty}^{infty} e^{-at} e^{st} dF(s) = e^{-at} mathbb{E} (e^{t sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $




      Now, I have really no clue on how to establish a bound at $c_{n}$ using the fact that $M_{X}(t) = 0 $ and $dM_{X}(t)/dt$ $vert_{t=0} = mathbb{E}(X)$ to show that if $mathbb{E}(X)<0$ then there is a $0>c<1$ with $P(S_{n}>a) leq c^{n}$. I have no attempt to show but any ideas would be very welcome.



      Thanks in advance






      statistics random-variables moment-generating-functions







      share|cite|improve this question









      $endgroup$




      I have the following question



      Let




      $X_{1}$, $X_{2}$, ..., $X_{n}$ be independent and identically
      distributed random variables with moment generating function
      $M_{X}(t)$, for -h < t < h, h > 0, $S_{n}$ = $sum_{i=1}^{n}X_i$, and
      $bar{X}$ = $S_{n}/n$.




      Ultimately I need to establish the following:




      Use the fact that $M_{X}(0)=1$ and $dM_{X}(t)/dt$ $vert_{t=0} =
      > mathbb{E}(X)$       to show that if $mathbb{E}(X)<0$ then there is a
      $0>c<1$ with $P(S_{n}>a) leq c^{n}$. Establish a similar bound for
      $P(S_{n} leq a)$.




      I have previously shown that




      $P(S_{n}>a) leq e^{-at}[M_{X}(t)]^{n}$ for $0<t<h$ , $ageq0$ and $P(S_{n}leq a) leq e^{-at}[M_{X}(t)]^{n}$ for $ -h<t<0 $




      As follows:




      $P(S_{n} > a) = int_a^{infty} dF(s) leq int_a^{infty} e^{-at} e^{st} dF(s) leq int_{-infty}^{infty} e^{-at} e^{st} dF(s) = e^{-at} mathbb{E} (e^{t sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $




      And similarly:




      $P(S_{n} leq a) = int_{-infty}^{a} dF(s) leq int_{-infty}^{a} e^{-at} e^{st} dF(s) leq int_{-infty}^{infty} e^{-at} e^{st} dF(s) = e^{-at} mathbb{E} (e^{t sum_{i=1}^{n}X_i}) = e^{-at}[M_{X}(t)]^{n} $




      Now, I have really no clue on how to establish a bound at $c_{n}$ using the fact that $M_{X}(t) = 0 $ and $dM_{X}(t)/dt$ $vert_{t=0} = mathbb{E}(X)$ to show that if $mathbb{E}(X)<0$ then there is a $0>c<1$ with $P(S_{n}>a) leq c^{n}$. I have no attempt to show but any ideas would be very welcome.



      Thanks in advance






      statistics random-variables moment-generating-functions




      statistics random-variables moment-generating-functions






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      asked Dec 13 '18 at 12:18









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