Intuition behind Expected Value of the Square of a Random Variable $X$
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 Nov 18 at 15:32
SABOY
549311
add a comment |
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 Nov 18 at 15:32
SABOY
549311
add a comment |
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 Nov 18 at 15:32
SABOY
549311
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 Nov 18 at 15:32
SABOY
549311
asked Nov 18 at 15:32
SABOY
549311
asked Nov 18 at 15:32
SABOY
549311
asked Nov 18 at 15:32
SABOY
549311
asked Nov 18 at 15:32
SABOY
549311
549311
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 Nov 18 at 16:05
drhab
97.8k544129
add a comment |
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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 Nov 18 at 16:05
drhab
97.8k544129
add a comment |
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 Nov 18 at 16:05
drhab
97.8k544129
add a comment |
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 Nov 18 at 16:05
drhab
97.8k544129
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 Nov 18 at 16:05
drhab
97.8k544129
edited Nov 19 at 7:46
answered Nov 18 at 16:05
drhab
97.8k544129
answered Nov 18 at 16:05
drhab
97.8k544129
answered Nov 18 at 16:05
drhab
97.8k544129
97.8k544129
add a comment |
add a comment |
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