Probability Distribution over a coomon finite alphabet
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Can anyone please give a hint about how to solve this problem? I donot how to start.
Thanks
probability probability-theory
asked Jan 12 at 11:22
Salwa MostafaSalwa Mostafa
323
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add a comment |
$begingroup$
Can anyone please give a hint about how to solve this problem? I donot how to start.
Thanks
probability probability-theory
asked Jan 12 at 11:22
Salwa MostafaSalwa Mostafa
323
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add a comment |
$begingroup$
Can anyone please give a hint about how to solve this problem? I donot how to start.
Thanks
probability probability-theory
asked Jan 12 at 11:22
Salwa MostafaSalwa Mostafa
323
$endgroup$
Can anyone please give a hint about how to solve this problem? I donot how to start.
Thanks
probability probability-theory
probability probability-theory
asked Jan 12 at 11:22
Salwa MostafaSalwa Mostafa
323
asked Jan 12 at 11:22
Salwa MostafaSalwa Mostafa
323
asked Jan 12 at 11:22
Salwa MostafaSalwa Mostafa
323
asked Jan 12 at 11:22
Salwa MostafaSalwa Mostafa
323
asked Jan 12 at 11:22
Salwa MostafaSalwa Mostafa
323
323
add a comment |
add a comment |
1 Answer
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Just let the alphabet $mathcal{A}$ to be, without loss of generality, ${1,2,dots,N}$. With this, you have,
$$
|p_k-p|_{TV}to 0 iff lim_{kto infty}sum_{j=1}^N |p_k(j)-p(j)| =0.
$$
That is, for every $jin[N]$, $|p_k(j)-p(j)|to 0$, as $kto infty$, thus, $|p_k(j)-p(j)|^2 to 0$, and thus, $sum_{j=1}^N |p_k(j)-p(j)|^2 to 0$, which says nothing but $p_kto p$ in $L^2$ sense.
answered Jan 12 at 17:43
AaronAaron
1,838415
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just let the alphabet $mathcal{A}$ to be, without loss of generality, ${1,2,dots,N}$. With this, you have,
$$
|p_k-p|_{TV}to 0 iff lim_{kto infty}sum_{j=1}^N |p_k(j)-p(j)| =0.
$$
That is, for every $jin[N]$, $|p_k(j)-p(j)|to 0$, as $kto infty$, thus, $|p_k(j)-p(j)|^2 to 0$, and thus, $sum_{j=1}^N |p_k(j)-p(j)|^2 to 0$, which says nothing but $p_kto p$ in $L^2$ sense.
answered Jan 12 at 17:43
AaronAaron
1,838415
$endgroup$
add a comment |
$begingroup$
Just let the alphabet $mathcal{A}$ to be, without loss of generality, ${1,2,dots,N}$. With this, you have,
$$
|p_k-p|_{TV}to 0 iff lim_{kto infty}sum_{j=1}^N |p_k(j)-p(j)| =0.
$$
That is, for every $jin[N]$, $|p_k(j)-p(j)|to 0$, as $kto infty$, thus, $|p_k(j)-p(j)|^2 to 0$, and thus, $sum_{j=1}^N |p_k(j)-p(j)|^2 to 0$, which says nothing but $p_kto p$ in $L^2$ sense.
answered Jan 12 at 17:43
AaronAaron
1,838415
$endgroup$
add a comment |
$begingroup$
Just let the alphabet $mathcal{A}$ to be, without loss of generality, ${1,2,dots,N}$. With this, you have,
$$
|p_k-p|_{TV}to 0 iff lim_{kto infty}sum_{j=1}^N |p_k(j)-p(j)| =0.
$$
That is, for every $jin[N]$, $|p_k(j)-p(j)|to 0$, as $kto infty$, thus, $|p_k(j)-p(j)|^2 to 0$, and thus, $sum_{j=1}^N |p_k(j)-p(j)|^2 to 0$, which says nothing but $p_kto p$ in $L^2$ sense.
answered Jan 12 at 17:43
AaronAaron
1,838415
$endgroup$
Just let the alphabet $mathcal{A}$ to be, without loss of generality, ${1,2,dots,N}$. With this, you have,
$$
|p_k-p|_{TV}to 0 iff lim_{kto infty}sum_{j=1}^N |p_k(j)-p(j)| =0.
$$
That is, for every $jin[N]$, $|p_k(j)-p(j)|to 0$, as $kto infty$, thus, $|p_k(j)-p(j)|^2 to 0$, and thus, $sum_{j=1}^N |p_k(j)-p(j)|^2 to 0$, which says nothing but $p_kto p$ in $L^2$ sense.
answered Jan 12 at 17:43
AaronAaron
1,838415
answered Jan 12 at 17:43
AaronAaron
1,838415
answered Jan 12 at 17:43
AaronAaron
1,838415
answered Jan 12 at 17:43
AaronAaron
1,838415
1,838415
add a comment |
add a comment |
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