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Intuition behind Expected Value of the Square of a Random Variable $X$
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Let's look at the case of $X sim Pois(lambda)$.
Since $k in X(Omega)$, it is clear that $k^2 in X(Omega)^{2}$. Following this logic and
from an intuitive view I'd say $mathbb E[X^{2}]=sum_{k^{2} in X(Omega)^{2}}k^{2}P(X^{2}=k^{2})$, however, I have seen that instead:
$mathbb E[X^{2}]=sum_{k in X({Omega})}k^{2}P(X=k)$. Can someone explain this notation to me, so that I can understand why the first mentioned is wrong and the latter correct.
probability-theory poisson-distribution expected-value
asked 22 hours ago
SABOY
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Let's look at the case of $X sim Pois(lambda)$.
Since $k in X(Omega)$, it is clear that $k^2 in X(Omega)^{2}$. Following this logic and
from an intuitive view I'd say $mathbb E[X^{2}]=sum_{k^{2} in X(Omega)^{2}}k^{2}P(X^{2}=k^{2})$, however, I have seen that instead:
$mathbb E[X^{2}]=sum_{k in X({Omega})}k^{2}P(X=k)$. Can someone explain this notation to me, so that I can understand why the first mentioned is wrong and the latter correct.
probability-theory poisson-distribution expected-value
asked 22 hours ago
SABOY
538211
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let's look at the case of $X sim Pois(lambda)$.
Since $k in X(Omega)$, it is clear that $k^2 in X(Omega)^{2}$. Following this logic and
from an intuitive view I'd say $mathbb E[X^{2}]=sum_{k^{2} in X(Omega)^{2}}k^{2}P(X^{2}=k^{2})$, however, I have seen that instead:
$mathbb E[X^{2}]=sum_{k in X({Omega})}k^{2}P(X=k)$. Can someone explain this notation to me, so that I can understand why the first mentioned is wrong and the latter correct.
probability-theory poisson-distribution expected-value
asked 22 hours ago
SABOY
538211
Let's look at the case of $X sim Pois(lambda)$.
Since $k in X(Omega)$, it is clear that $k^2 in X(Omega)^{2}$. Following this logic and
from an intuitive view I'd say $mathbb E[X^{2}]=sum_{k^{2} in X(Omega)^{2}}k^{2}P(X^{2}=k^{2})$, however, I have seen that instead:
$mathbb E[X^{2}]=sum_{k in X({Omega})}k^{2}P(X=k)$. Can someone explain this notation to me, so that I can understand why the first mentioned is wrong and the latter correct.
probability-theory poisson-distribution expected-value
probability-theory poisson-distribution expected-value
asked 22 hours ago
SABOY
538211
asked 22 hours ago
SABOY
538211
asked 22 hours ago
SABOY
538211
asked 22 hours ago
SABOY
538211
asked 22 hours ago
SABOY
538211
538211
add a comment |
add a comment |
1 Answer
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If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.
If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.
Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.
It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.
(You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)
Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.
The same objection counts for the use of $X(Omega)$.
If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$
Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.
answered 22 hours ago
drhab
94.2k543125
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.
If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.
Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.
It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.
(You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)
Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.
The same objection counts for the use of $X(Omega)$.
If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$
Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.
answered 22 hours ago
drhab
94.2k543125
add a comment |
up vote
1
down vote
accepted
If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.
If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.
Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.
It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.
(You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)
Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.
The same objection counts for the use of $X(Omega)$.
If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$
Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.
answered 22 hours ago
drhab
94.2k543125
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.
If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.
Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.
It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.
(You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)
Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.
The same objection counts for the use of $X(Omega)$.
If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$
Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.
answered 22 hours ago
drhab
94.2k543125
If $X(Omega)$ is uncountable then both notations are wrong because uncountable sums are not defined.
If $X(Omega)$ is countable then both notations cannot be classified as "wrong" but as tending to redundancy, and both notations should not be used.
Note that here $P(X^2=k^2)=P(X=k)$ so that the summations will have the same outcome.
It is tricky to write $X(Omega)^2$ because this can also be a notation for the set $X(Omega)times X(Omega)$.
(You most probably want it to be defined as ${X(omega)^2midomegainOmega}$.)
Further it is somewhat redundant to use $X(Omega)^2$ as a set over which $k^2$ ranges, because it is clear allready that for every $k^2notin X(Omega)^2$ evidently $P(X^2=k^2)=0$ so that the term on forehand is excluded.
The same objection counts for the use of $X(Omega)$.
If a random variable is discrete then we always can find a countable set $Ssubseteq mathbb R$ with $P(Xin S)=1$ and for notation of expectations we can practicize the more handsome notation: $$mathbb Ef(X)=sum_{kin S}f(k)P(X=k)$$
Also it is possible to find a minimal $S$ that does that job, and we can identify this $S$ as the support of the discrete random variable.
answered 22 hours ago
drhab
94.2k543125
edited 6 hours ago
answered 22 hours ago
drhab
94.2k543125
answered 22 hours ago
drhab
94.2k543125
answered 22 hours ago
drhab
94.2k543125
94.2k543125
add a comment |
add a comment |
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