Suppose $lim_{n to infty}a_n=a$, Prove $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.
$begingroup$
Suppose $lim_{n to infty}a_n=a$. Prove that $$lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a,.$$
Here is a proof,
Proof: Consider $t_{n,k}=frac{1}{2^n}binom{n}{k}$,$0 leq t_{n,k}= frac{n(n-1)...(n-k+1)}{2^nk(k-1)...1}leqfrac{n^k}{2^n} $ and by squeeze theorem, $lim_{n to infty}t_{n,k}=0$. Also, $sum_{k=0}^n frac{1}{2^n}binom{n}{k}=1$, so then by Silverman–Toeplitz theorem, we have $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.
Is this proof correct? Is there other cool proofs?
real-analysis sequences-and-series limits proof-verification alternative-proof
$endgroup$
add a comment |
$begingroup$
Suppose $lim_{n to infty}a_n=a$. Prove that $$lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a,.$$
Here is a proof,
Proof: Consider $t_{n,k}=frac{1}{2^n}binom{n}{k}$,$0 leq t_{n,k}= frac{n(n-1)...(n-k+1)}{2^nk(k-1)...1}leqfrac{n^k}{2^n} $ and by squeeze theorem, $lim_{n to infty}t_{n,k}=0$. Also, $sum_{k=0}^n frac{1}{2^n}binom{n}{k}=1$, so then by Silverman–Toeplitz theorem, we have $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.
Is this proof correct? Is there other cool proofs?
real-analysis sequences-and-series limits proof-verification alternative-proof
$endgroup$
1
$begingroup$
What is topliz?
$endgroup$
– Batominovski
Dec 14 '18 at 3:41
1
$begingroup$
Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
$endgroup$
– T. Bongers
Dec 14 '18 at 3:46
$begingroup$
Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
$endgroup$
– nafhgood
Dec 14 '18 at 3:46
$begingroup$
I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
$endgroup$
– Robert Wolfe
Dec 14 '18 at 5:58
add a comment |
$begingroup$
Suppose $lim_{n to infty}a_n=a$. Prove that $$lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a,.$$
Here is a proof,
Proof: Consider $t_{n,k}=frac{1}{2^n}binom{n}{k}$,$0 leq t_{n,k}= frac{n(n-1)...(n-k+1)}{2^nk(k-1)...1}leqfrac{n^k}{2^n} $ and by squeeze theorem, $lim_{n to infty}t_{n,k}=0$. Also, $sum_{k=0}^n frac{1}{2^n}binom{n}{k}=1$, so then by Silverman–Toeplitz theorem, we have $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.
Is this proof correct? Is there other cool proofs?
real-analysis sequences-and-series limits proof-verification alternative-proof
$endgroup$
Suppose $lim_{n to infty}a_n=a$. Prove that $$lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a,.$$
Here is a proof,
Proof: Consider $t_{n,k}=frac{1}{2^n}binom{n}{k}$,$0 leq t_{n,k}= frac{n(n-1)...(n-k+1)}{2^nk(k-1)...1}leqfrac{n^k}{2^n} $ and by squeeze theorem, $lim_{n to infty}t_{n,k}=0$. Also, $sum_{k=0}^n frac{1}{2^n}binom{n}{k}=1$, so then by Silverman–Toeplitz theorem, we have $lim_{nto infty} frac{1}{2^n} sum_{k=0}^n binom{n}{k}a_k=a$.
Is this proof correct? Is there other cool proofs?
real-analysis sequences-and-series limits proof-verification alternative-proof
real-analysis sequences-and-series limits proof-verification alternative-proof
edited Dec 14 '18 at 3:45
nafhgood
asked Dec 14 '18 at 3:33
nafhgoodnafhgood
1,805422
1,805422
1
$begingroup$
What is topliz?
$endgroup$
– Batominovski
Dec 14 '18 at 3:41
1
$begingroup$
Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
$endgroup$
– T. Bongers
Dec 14 '18 at 3:46
$begingroup$
Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
$endgroup$
– nafhgood
Dec 14 '18 at 3:46
$begingroup$
I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
$endgroup$
– Robert Wolfe
Dec 14 '18 at 5:58
add a comment |
1
$begingroup$
What is topliz?
$endgroup$
– Batominovski
Dec 14 '18 at 3:41
1
$begingroup$
Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
$endgroup$
– T. Bongers
Dec 14 '18 at 3:46
$begingroup$
Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
$endgroup$
– nafhgood
Dec 14 '18 at 3:46
$begingroup$
I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
$endgroup$
– Robert Wolfe
Dec 14 '18 at 5:58
1
1
$begingroup$
What is topliz?
$endgroup$
– Batominovski
Dec 14 '18 at 3:41
$begingroup$
What is topliz?
$endgroup$
– Batominovski
Dec 14 '18 at 3:41
1
1
$begingroup$
Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
$endgroup$
– T. Bongers
Dec 14 '18 at 3:46
$begingroup$
Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
$endgroup$
– T. Bongers
Dec 14 '18 at 3:46
$begingroup$
Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
$endgroup$
– nafhgood
Dec 14 '18 at 3:46
$begingroup$
Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
$endgroup$
– nafhgood
Dec 14 '18 at 3:46
$begingroup$
I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
$endgroup$
– Robert Wolfe
Dec 14 '18 at 5:58
$begingroup$
I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
$endgroup$
– Robert Wolfe
Dec 14 '18 at 5:58
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.
$endgroup$
add a comment |
$begingroup$
There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
$$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
It's reasonable to suspect that a similar approach there might work here.
After one starts down this road, you eventually realize that you'll need to show that
$$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$
where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
$$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
at least once $n$ surpasses $N$. This inequality gives the limit we wanted.
Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
$$|a_n-a|<frac{epsilon}{2}$$
for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
$$|a_n-a|<M,.$$
Choose $N_2$ such that
$$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:
begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
&=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
&leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
end{align*}
for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
$$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.
$endgroup$
add a comment |
$begingroup$
Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.
$endgroup$
add a comment |
$begingroup$
Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.
$endgroup$
Say $|a_n| le C$ for each $n$. For $epsilon > 0$, take $N$ s.t. $|a_n-a| < epsilon$ for $n ge N$. Then for $M ge N^2$, $$frac{1}{2^M}sum_{k=0}^M {M choose k}|a_k-a| le frac{1}{2^M}sum_{k=0}^sqrt{M} {M choose k}2C+frac{1}{2^M}sum_{k=sqrt{M}}^M {M choose k}epsilon.$$ So, $limsup_{M to infty} |frac{1}{2^M}sum_{k=0}^M {M choose k} a_k - a| le epsilon$. This holds for each $epsilon > 0$.
answered Dec 14 '18 at 3:41
mathworker21mathworker21
8,9421928
8,9421928
add a comment |
add a comment |
$begingroup$
There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
$$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
It's reasonable to suspect that a similar approach there might work here.
After one starts down this road, you eventually realize that you'll need to show that
$$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$
where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
$$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
at least once $n$ surpasses $N$. This inequality gives the limit we wanted.
Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
$$|a_n-a|<frac{epsilon}{2}$$
for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
$$|a_n-a|<M,.$$
Choose $N_2$ such that
$$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:
begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
&=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
&leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
end{align*}
for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
$$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$
$endgroup$
add a comment |
$begingroup$
There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
$$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
It's reasonable to suspect that a similar approach there might work here.
After one starts down this road, you eventually realize that you'll need to show that
$$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$
where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
$$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
at least once $n$ surpasses $N$. This inequality gives the limit we wanted.
Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
$$|a_n-a|<frac{epsilon}{2}$$
for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
$$|a_n-a|<M,.$$
Choose $N_2$ such that
$$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:
begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
&=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
&leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
end{align*}
for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
$$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$
$endgroup$
add a comment |
$begingroup$
There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
$$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
It's reasonable to suspect that a similar approach there might work here.
After one starts down this road, you eventually realize that you'll need to show that
$$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$
where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
$$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
at least once $n$ surpasses $N$. This inequality gives the limit we wanted.
Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
$$|a_n-a|<frac{epsilon}{2}$$
for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
$$|a_n-a|<M,.$$
Choose $N_2$ such that
$$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:
begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
&=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
&leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
end{align*}
for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
$$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$
$endgroup$
There's a pretty standard exercise in which we are asked to show that if $a_krightarrow a$ then
$$biggl(frac{1}{n}sum_{k=1}^n a_kbiggr)rightarrow a,.$$
It's reasonable to suspect that a similar approach there might work here.
After one starts down this road, you eventually realize that you'll need to show that
$$lim_{nrightarrowinfty}frac{1}{2^n}sum_{k=0}^Nbinom{n}{k}=0$$
where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice
$$sum_{k=0}^Nbinom{n}{k}leqsum_{k=0}^Nfrac{n^k}{k!}=sum_{k=0}^Nfrac{N^k}{k!}left(frac{n}{N}right)^kleqleft(frac{n}{N}right)^Ne^N$$
at least once $n$ surpasses $N$. This inequality gives the limit we wanted.
Now we are ready to start the formal presentation. Let $epsilon>0$. Choose $N_1$ in $mathbb{N}$ such that
$$|a_n-a|<frac{epsilon}{2}$$
for every $ngeq N_1$. Let $M$ satisfy the inequality below for every $n$ in $mathbb{N}$
$$|a_n-a|<M,.$$
Choose $N_2$ such that
$$frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}<frac{epsilon}{2(M+1)}$$
for every $ngeq N_2$. We now observe the scratch-paper inequalities that inspired the approach:
begin{align*}left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|
&=left|sum_{k=0}^nfrac{1}{2^n}binom{n}{k}(a_k-a)right|\
&leq frac{1}{2^n}sum_{k=0}^{N_1-1}binom{n}{k}|a_k-a|+sum_{k=N_1}^nfrac{1}{2^n}binom{n}{k}|a_k-a|
end{align*}
for sufficiently large $n$. In particular, if we take $n>max(N_1, N_2)$, we then get
$$left|biggl(,sum_{k=0}^nfrac{1}{2^n}binom{n}{k}a_kbiggr)-a,right|<epsilon,.$$
answered Dec 14 '18 at 14:35
Robert WolfeRobert Wolfe
5,79722763
5,79722763
add a comment |
add a comment |
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What is topliz?
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– Batominovski
Dec 14 '18 at 3:41
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Ultimately this is just a consequence of the fact that $$frac 1 {2^n} sum_{k = 0}^n binom{n}{k} = 1$$ while the individual summands tend to zero, so this acts as an averaging operator. You may be interested in the ergodic theorem, which considers a similar class of averages.
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– T. Bongers
Dec 14 '18 at 3:46
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Sorry! It is Silverman–Toeplitz theorem, I miss spelled it :en.wikipedia.org/wiki/Silverman%E2%80%93Toeplitz_theorem
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– nafhgood
Dec 14 '18 at 3:46
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I'm sure many probabilistic proofs could be given given that this question could be cast as a sequence of random variables. You might draw in that attention with the correct tag.
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– Robert Wolfe
Dec 14 '18 at 5:58