Geometric distribution of independent t random variables and their limit
$begingroup$
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
asked Jan 5 at 17:33
LonelyStudentLonelyStudent
523
$endgroup$
add a comment |
$begingroup$
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
asked Jan 5 at 17:33
LonelyStudentLonelyStudent
523
$endgroup$
$begingroup$
Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
$endgroup$
– Mike Earnest
Jan 5 at 18:28
$begingroup$
Yes you are right, i edited.
$endgroup$
– LonelyStudent
Jan 5 at 18:29
add a comment |
$begingroup$
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
asked Jan 5 at 17:33
LonelyStudentLonelyStudent
523
$endgroup$
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
asked Jan 5 at 17:33
LonelyStudentLonelyStudent
523
asked Jan 5 at 17:33
LonelyStudentLonelyStudent
523
edited Jan 5 at 18:30
LonelyStudent
asked Jan 5 at 17:33
LonelyStudentLonelyStudent
523
asked Jan 5 at 17:33
LonelyStudentLonelyStudent
523
asked Jan 5 at 17:33
LonelyStudentLonelyStudent
523
523
$begingroup$
Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
$endgroup$
– Mike Earnest
Jan 5 at 18:28
$begingroup$
Yes you are right, i edited.
$endgroup$
– LonelyStudent
Jan 5 at 18:29
add a comment |
$begingroup$
Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
$endgroup$
– Mike Earnest
Jan 5 at 18:28
$begingroup$
Yes you are right, i edited.
$endgroup$
– LonelyStudent
Jan 5 at 18:29
$begingroup$
Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
$endgroup$
– Mike Earnest
Jan 5 at 18:28
$begingroup$
Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
$endgroup$
– Mike Earnest
Jan 5 at 18:28
$begingroup$
Yes you are right, i edited.
$endgroup$
– LonelyStudent
Jan 5 at 18:29
$begingroup$
Yes you are right, i edited.
$endgroup$
– LonelyStudent
Jan 5 at 18:29
add a comment |
1 Answer
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$begingroup$
$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
$$
P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$
answered Jan 5 at 18:35
Mike EarnestMike Earnest
25.2k22151
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add a comment |
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1 Answer
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active
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1 Answer
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$begingroup$
$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
$$
P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$
answered Jan 5 at 18:35
Mike EarnestMike Earnest
25.2k22151
$endgroup$
add a comment |
$begingroup$
$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
$$
P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$
answered Jan 5 at 18:35
Mike EarnestMike Earnest
25.2k22151
$endgroup$
add a comment |
$begingroup$
$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
$$
P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$
answered Jan 5 at 18:35
Mike EarnestMike Earnest
25.2k22151
$endgroup$
$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
$$
P(X_1+dots+X_tge at)le e^{-sat}cdot e^{t/2}=e^{-(sa-frac12)t}ge e^{-(sa-frac23sa)t}=2^{-(log_2 e)sat/3}.$$
answered Jan 5 at 18:35
Mike EarnestMike Earnest
25.2k22151
edited Jan 7 at 16:50
answered Jan 5 at 18:35
Mike EarnestMike Earnest
25.2k22151
answered Jan 5 at 18:35
Mike EarnestMike Earnest
25.2k22151
answered Jan 5 at 18:35
Mike EarnestMike Earnest
25.2k22151
25.2k22151
add a comment |
add a comment |
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$begingroup$
Are you sure you don't mean $2^{-cat}$? $2^{cat}$ is bigger than $1$, so the probability is trivially less than $2^{cat}$.
$endgroup$
– Mike Earnest
Jan 5 at 18:28
$begingroup$
Yes you are right, i edited.
$endgroup$
– LonelyStudent
Jan 5 at 18:29