Prove that almost surely $limsup_{ntoinfty}frac{X_n}{ln n}=frac{1}{lambda}$
$begingroup$
Let ${X_n}$, n=1 to infinity, be independent random variables distributed $Exp(lambda)$.
Prove that almost surely $$limsup_{ntoinfty}frac{X_n}{ln n}=frac{1}{lambda}$$
My idea was to look at the $$sum_{k=1}^{infty}Pleft(left|frac{X_k}{ln k}-frac{1}{lambda}right|>epsilonright)$$ and prove it is finite. But I think it is not.
$$Pleft(X_k<left(-epsilon+frac{1}{lambda}right)ln kright)=frac{1}{k}(1-k^{lambdaepsilon})$$ and we can take such espilon value so it is divergent as $ktoinfty$, which means sum does not converge.
probability probability-distributions convergence exponential-distribution almost-everywhere
edited Dec 5 '18 at 14:06
the_candyman
8,84622045
asked Dec 5 '18 at 14:02
ryszard egginkryszard eggink
315110
$endgroup$
add a comment |
$begingroup$
Let ${X_n}$, n=1 to infinity, be independent random variables distributed $Exp(lambda)$.
Prove that almost surely $$limsup_{ntoinfty}frac{X_n}{ln n}=frac{1}{lambda}$$
My idea was to look at the $$sum_{k=1}^{infty}Pleft(left|frac{X_k}{ln k}-frac{1}{lambda}right|>epsilonright)$$ and prove it is finite. But I think it is not.
$$Pleft(X_k<left(-epsilon+frac{1}{lambda}right)ln kright)=frac{1}{k}(1-k^{lambdaepsilon})$$ and we can take such espilon value so it is divergent as $ktoinfty$, which means sum does not converge.
probability probability-distributions convergence exponential-distribution almost-everywhere
edited Dec 5 '18 at 14:06
the_candyman
8,84622045
asked Dec 5 '18 at 14:02
ryszard egginkryszard eggink
315110
$endgroup$
add a comment |
$begingroup$
Let ${X_n}$, n=1 to infinity, be independent random variables distributed $Exp(lambda)$.
Prove that almost surely $$limsup_{ntoinfty}frac{X_n}{ln n}=frac{1}{lambda}$$
My idea was to look at the $$sum_{k=1}^{infty}Pleft(left|frac{X_k}{ln k}-frac{1}{lambda}right|>epsilonright)$$ and prove it is finite. But I think it is not.
$$Pleft(X_k<left(-epsilon+frac{1}{lambda}right)ln kright)=frac{1}{k}(1-k^{lambdaepsilon})$$ and we can take such espilon value so it is divergent as $ktoinfty$, which means sum does not converge.
probability probability-distributions convergence exponential-distribution almost-everywhere
edited Dec 5 '18 at 14:06
the_candyman
8,84622045
asked Dec 5 '18 at 14:02
ryszard egginkryszard eggink
315110
$endgroup$
Let ${X_n}$, n=1 to infinity, be independent random variables distributed $Exp(lambda)$.
Prove that almost surely $$limsup_{ntoinfty}frac{X_n}{ln n}=frac{1}{lambda}$$
My idea was to look at the $$sum_{k=1}^{infty}Pleft(left|frac{X_k}{ln k}-frac{1}{lambda}right|>epsilonright)$$ and prove it is finite. But I think it is not.
$$Pleft(X_k<left(-epsilon+frac{1}{lambda}right)ln kright)=frac{1}{k}(1-k^{lambdaepsilon})$$ and we can take such espilon value so it is divergent as $ktoinfty$, which means sum does not converge.
probability probability-distributions convergence exponential-distribution almost-everywhere
probability probability-distributions convergence exponential-distribution almost-everywhere
edited Dec 5 '18 at 14:06
the_candyman
8,84622045
asked Dec 5 '18 at 14:02
ryszard egginkryszard eggink
315110
edited Dec 5 '18 at 14:06
the_candyman
8,84622045
asked Dec 5 '18 at 14:02
ryszard egginkryszard eggink
315110
edited Dec 5 '18 at 14:06
the_candyman
8,84622045
edited Dec 5 '18 at 14:06
the_candyman
8,84622045
edited Dec 5 '18 at 14:06
the_candyman
8,84622045
8,84622045
asked Dec 5 '18 at 14:02
ryszard egginkryszard eggink
315110
asked Dec 5 '18 at 14:02
ryszard egginkryszard eggink
315110
asked Dec 5 '18 at 14:02
ryszard egginkryszard eggink
315110
315110
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Observe that for $c>0$, we have that
$$
sum_{n=0}^infty P(X_n>clog n)=sum_{n=0}^inftyfrac{1}{n^{lambda c}}=begin{cases}
<infty&text{if }c>1/lambda\
infty&text{if }c<1/lambda
end{cases}quad (1)
$$
where we used the fact that $P(X>x)=e^{-lambda x}$ for $xgeq 0$ if $Xsim text{exp}(lambda)$. It follows by Borel-Cantelli that for each $varepsilon>0$
$$
Pleft(frac{X_n}{log n}>lambda^{-1}+varepsilonquad text{i.o}right)=0
$$
whence
$$
limsup frac{X_n}{log n}leq lambda^{-1}.
$$
with probability $1$. Similarly for each $varepsilon>0$,
$$
Pleft(frac{X_n}{log n}>lambda^{-1}-varepsilonquad text{i.o}right)=1
$$
whence
$$
limsup frac{X_n}{log n}geq lambda^{-1}.$$
with probability $1$.
answered Dec 5 '18 at 14:22
Foobaz JohnFoobaz John
21.8k41352
$endgroup$
$begingroup$
thanks! :) nice work
$endgroup$
– ryszard eggink
Dec 5 '18 at 14:26
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Observe that for $c>0$, we have that
$$
sum_{n=0}^infty P(X_n>clog n)=sum_{n=0}^inftyfrac{1}{n^{lambda c}}=begin{cases}
<infty&text{if }c>1/lambda\
infty&text{if }c<1/lambda
end{cases}quad (1)
$$
where we used the fact that $P(X>x)=e^{-lambda x}$ for $xgeq 0$ if $Xsim text{exp}(lambda)$. It follows by Borel-Cantelli that for each $varepsilon>0$
$$
Pleft(frac{X_n}{log n}>lambda^{-1}+varepsilonquad text{i.o}right)=0
$$
whence
$$
limsup frac{X_n}{log n}leq lambda^{-1}.
$$
with probability $1$. Similarly for each $varepsilon>0$,
$$
Pleft(frac{X_n}{log n}>lambda^{-1}-varepsilonquad text{i.o}right)=1
$$
whence
$$
limsup frac{X_n}{log n}geq lambda^{-1}.$$
with probability $1$.
answered Dec 5 '18 at 14:22
Foobaz JohnFoobaz John
21.8k41352
$endgroup$
$begingroup$
thanks! :) nice work
$endgroup$
– ryszard eggink
Dec 5 '18 at 14:26
add a comment |
$begingroup$
Observe that for $c>0$, we have that
$$
sum_{n=0}^infty P(X_n>clog n)=sum_{n=0}^inftyfrac{1}{n^{lambda c}}=begin{cases}
<infty&text{if }c>1/lambda\
infty&text{if }c<1/lambda
end{cases}quad (1)
$$
where we used the fact that $P(X>x)=e^{-lambda x}$ for $xgeq 0$ if $Xsim text{exp}(lambda)$. It follows by Borel-Cantelli that for each $varepsilon>0$
$$
Pleft(frac{X_n}{log n}>lambda^{-1}+varepsilonquad text{i.o}right)=0
$$
whence
$$
limsup frac{X_n}{log n}leq lambda^{-1}.
$$
with probability $1$. Similarly for each $varepsilon>0$,
$$
Pleft(frac{X_n}{log n}>lambda^{-1}-varepsilonquad text{i.o}right)=1
$$
whence
$$
limsup frac{X_n}{log n}geq lambda^{-1}.$$
with probability $1$.
answered Dec 5 '18 at 14:22
Foobaz JohnFoobaz John
21.8k41352
$endgroup$
$begingroup$
thanks! :) nice work
$endgroup$
– ryszard eggink
Dec 5 '18 at 14:26
add a comment |
$begingroup$
Observe that for $c>0$, we have that
$$
sum_{n=0}^infty P(X_n>clog n)=sum_{n=0}^inftyfrac{1}{n^{lambda c}}=begin{cases}
<infty&text{if }c>1/lambda\
infty&text{if }c<1/lambda
end{cases}quad (1)
$$
where we used the fact that $P(X>x)=e^{-lambda x}$ for $xgeq 0$ if $Xsim text{exp}(lambda)$. It follows by Borel-Cantelli that for each $varepsilon>0$
$$
Pleft(frac{X_n}{log n}>lambda^{-1}+varepsilonquad text{i.o}right)=0
$$
whence
$$
limsup frac{X_n}{log n}leq lambda^{-1}.
$$
with probability $1$. Similarly for each $varepsilon>0$,
$$
Pleft(frac{X_n}{log n}>lambda^{-1}-varepsilonquad text{i.o}right)=1
$$
whence
$$
limsup frac{X_n}{log n}geq lambda^{-1}.$$
with probability $1$.
answered Dec 5 '18 at 14:22
Foobaz JohnFoobaz John
21.8k41352
$endgroup$
Observe that for $c>0$, we have that
$$
sum_{n=0}^infty P(X_n>clog n)=sum_{n=0}^inftyfrac{1}{n^{lambda c}}=begin{cases}
<infty&text{if }c>1/lambda\
infty&text{if }c<1/lambda
end{cases}quad (1)
$$
where we used the fact that $P(X>x)=e^{-lambda x}$ for $xgeq 0$ if $Xsim text{exp}(lambda)$. It follows by Borel-Cantelli that for each $varepsilon>0$
$$
Pleft(frac{X_n}{log n}>lambda^{-1}+varepsilonquad text{i.o}right)=0
$$
whence
$$
limsup frac{X_n}{log n}leq lambda^{-1}.
$$
with probability $1$. Similarly for each $varepsilon>0$,
$$
Pleft(frac{X_n}{log n}>lambda^{-1}-varepsilonquad text{i.o}right)=1
$$
whence
$$
limsup frac{X_n}{log n}geq lambda^{-1}.$$
with probability $1$.
answered Dec 5 '18 at 14:22
Foobaz JohnFoobaz John
21.8k41352
answered Dec 5 '18 at 14:22
Foobaz JohnFoobaz John
21.8k41352
answered Dec 5 '18 at 14:22
Foobaz JohnFoobaz John
21.8k41352
answered Dec 5 '18 at 14:22
Foobaz JohnFoobaz John
21.8k41352
21.8k41352
$begingroup$
thanks! :) nice work
$endgroup$
– ryszard eggink
Dec 5 '18 at 14:26
add a comment |
$begingroup$
thanks! :) nice work
$endgroup$
– ryszard eggink
Dec 5 '18 at 14:26
$begingroup$
thanks! :) nice work
$endgroup$
– ryszard eggink
Dec 5 '18 at 14:26
$begingroup$
thanks! :) nice work
$endgroup$
– ryszard eggink
Dec 5 '18 at 14:26
add a comment |
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