Geometric distribution of independent t random variables and their limit












2














let there be $X_1,X_2,...,X_t$ independent random variables that are distributed $X_itext{~}Geo(frac{1}{2})$ for every $1le i le t$. Show that there is a constant $c gt 0 $ so that for every $a gt 3$ and every $t ge 1$:



$$ P(X_1 + X_2 + ... + X_t ge a cdot t) le 2^{-c cdot a cdot t}$$



Mention specifically the constant and prove its correctness.



So far I tried using Chernoff bounds but I am not able to complete the calculation due to the $delta$, the expected value can be found due to the geometry of the the random variable, moreover is that they are independent so we can also find variance but I don't see a use for it at the moment.
I have also tried maybe getting the expression to the central limit theorem but that was a dead end as well, any help would be much appreciated.






probability







share|cite|improve this question
























  • Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
    – Mike Earnest
    2 days ago










  • Yes you are right, i edited.
    – LonelyStudent
    2 days ago
















2














let there be $X_1,X_2,...,X_t$ independent random variables that are distributed $X_itext{~}Geo(frac{1}{2})$ for every $1le i le t$. Show that there is a constant $c gt 0 $ so that for every $a gt 3$ and every $t ge 1$:



$$ P(X_1 + X_2 + ... + X_t ge a cdot t) le 2^{-c cdot a cdot t}$$



Mention specifically the constant and prove its correctness.



So far I tried using Chernoff bounds but I am not able to complete the calculation due to the $delta$, the expected value can be found due to the geometry of the the random variable, moreover is that they are independent so we can also find variance but I don't see a use for it at the moment.
I have also tried maybe getting the expression to the central limit theorem but that was a dead end as well, any help would be much appreciated.






probability







share|cite|improve this question
























  • Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
    – Mike Earnest
    2 days ago










  • Yes you are right, i edited.
    – LonelyStudent
    2 days ago














2












2








2







let there be $X_1,X_2,...,X_t$ independent random variables that are distributed $X_itext{~}Geo(frac{1}{2})$ for every $1le i le t$. Show that there is a constant $c gt 0 $ so that for every $a gt 3$ and every $t ge 1$:



$$ P(X_1 + X_2 + ... + X_t ge a cdot t) le 2^{-c cdot a cdot t}$$



Mention specifically the constant and prove its correctness.



So far I tried using Chernoff bounds but I am not able to complete the calculation due to the $delta$, the expected value can be found due to the geometry of the the random variable, moreover is that they are independent so we can also find variance but I don't see a use for it at the moment.
I have also tried maybe getting the expression to the central limit theorem but that was a dead end as well, any help would be much appreciated.






probability







share|cite|improve this question















let there be $X_1,X_2,...,X_t$ independent random variables that are distributed $X_itext{~}Geo(frac{1}{2})$ for every $1le i le t$. Show that there is a constant $c gt 0 $ so that for every $a gt 3$ and every $t ge 1$:



$$ P(X_1 + X_2 + ... + X_t ge a cdot t) le 2^{-c cdot a cdot t}$$



Mention specifically the constant and prove its correctness.



So far I tried using Chernoff bounds but I am not able to complete the calculation due to the $delta$, the expected value can be found due to the geometry of the the random variable, moreover is that they are independent so we can also find variance but I don't see a use for it at the moment.
I have also tried maybe getting the expression to the central limit theorem but that was a dead end as well, any help would be much appreciated.






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







LonelyStudent

















asked 2 days ago









LonelyStudentLonelyStudent

523




523












  • Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
    – Mike Earnest
    2 days ago










  • Yes you are right, i edited.
    – LonelyStudent
    2 days ago


















  • Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
    – Mike Earnest
    2 days ago










  • Yes you are right, i edited.
    – LonelyStudent
    2 days ago
















Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
– Mike Earnest
2 days ago




Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
– Mike Earnest
2 days ago












Yes you are right, i edited.
– LonelyStudent
2 days ago




Yes you are right, i edited.
– LonelyStudent
2 days ago










1 Answer
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$E[e^{sX_1}]=sum_k (1/2)^{k+1}e^{sk}=1/(2-e^s). $ Therefore, the mgf for $X_1+dots+X_t$ is $(2-e^{s})^{-t}$, valid for all $s<ln 2$.



For any $0<s<ln 2$,
$$
P(X_1+dots+X_tge at)=P(e^{s(X_1+dots+X_t)}ge e^{sat})lefrac{E[e^{s(X_1+dots+X_t)}]}{e^{sat}}= e^{-sat}(2-e^s)^{-t}.
$$
Now choose $s$ so that $2-e^s=e^{-1/2}$, namely $s=log(2-e^{-1/2})approx 0.33$. Then since $sa ge 0.33cdot 3ge 3/4$, we have $frac23 sage frac12$, so
$$
P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$






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    1 Answer
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    1 Answer
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    $E[e^{sX_1}]=sum_k (1/2)^{k+1}e^{sk}=1/(2-e^s). $ Therefore, the mgf for $X_1+dots+X_t$ is $(2-e^{s})^{-t}$, valid for all $s<ln 2$.



    For any $0<s<ln 2$,
    $$
    P(X_1+dots+X_tge at)=P(e^{s(X_1+dots+X_t)}ge e^{sat})lefrac{E[e^{s(X_1+dots+X_t)}]}{e^{sat}}= e^{-sat}(2-e^s)^{-t}.
    $$
    Now choose $s$ so that $2-e^s=e^{-1/2}$, namely $s=log(2-e^{-1/2})approx 0.33$. Then since $sa ge 0.33cdot 3ge 3/4$, we have $frac23 sage frac12$, so
    $$
    P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$






    share|cite|improve this answer




























      0














      $E[e^{sX_1}]=sum_k (1/2)^{k+1}e^{sk}=1/(2-e^s). $ Therefore, the mgf for $X_1+dots+X_t$ is $(2-e^{s})^{-t}$, valid for all $s<ln 2$.



      For any $0<s<ln 2$,
      $$
      P(X_1+dots+X_tge at)=P(e^{s(X_1+dots+X_t)}ge e^{sat})lefrac{E[e^{s(X_1+dots+X_t)}]}{e^{sat}}= e^{-sat}(2-e^s)^{-t}.
      $$
      Now choose $s$ so that $2-e^s=e^{-1/2}$, namely $s=log(2-e^{-1/2})approx 0.33$. Then since $sa ge 0.33cdot 3ge 3/4$, we have $frac23 sage frac12$, so
      $$
      P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$






      share|cite|improve this answer


























        0












        0








        0






        $E[e^{sX_1}]=sum_k (1/2)^{k+1}e^{sk}=1/(2-e^s). $ Therefore, the mgf for $X_1+dots+X_t$ is $(2-e^{s})^{-t}$, valid for all $s<ln 2$.



        For any $0<s<ln 2$,
        $$
        P(X_1+dots+X_tge at)=P(e^{s(X_1+dots+X_t)}ge e^{sat})lefrac{E[e^{s(X_1+dots+X_t)}]}{e^{sat}}= e^{-sat}(2-e^s)^{-t}.
        $$
        Now choose $s$ so that $2-e^s=e^{-1/2}$, namely $s=log(2-e^{-1/2})approx 0.33$. Then since $sa ge 0.33cdot 3ge 3/4$, we have $frac23 sage frac12$, so
        $$
        P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$






        share|cite|improve this answer














        $E[e^{sX_1}]=sum_k (1/2)^{k+1}e^{sk}=1/(2-e^s). $ Therefore, the mgf for $X_1+dots+X_t$ is $(2-e^{s})^{-t}$, valid for all $s<ln 2$.



        For any $0<s<ln 2$,
        $$
        P(X_1+dots+X_tge at)=P(e^{s(X_1+dots+X_t)}ge e^{sat})lefrac{E[e^{s(X_1+dots+X_t)}]}{e^{sat}}= e^{-sat}(2-e^s)^{-t}.
        $$
        Now choose $s$ so that $2-e^s=e^{-1/2}$, namely $s=log(2-e^{-1/2})approx 0.33$. Then since $sa ge 0.33cdot 3ge 3/4$, we have $frac23 sage frac12$, so
        $$
        P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 16 hours ago

























        answered 2 days ago









        Mike EarnestMike Earnest

        20.6k11950




        20.6k11950






























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