“Popularity Principle” in Probability Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Does this probability distribution have a name?A, B, C and D, and we are trying to find the probability that exactly one event occurs.Joint probability of two random variables both following poisson distributionExpectation of of joint probability distributionApplication of the pigeon hole principleTransformation of a non one-to-one probability mass function?A question in probability and combinatoricsProblem with my intuition concerning probabilities of dependent events and conditional probabilityDuel question, why shoot when $P_1 (s)+P_2 (s)=1$High F-Statistic Value and negative Probability for Granger Causality Result, Interpretation?
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“Popularity Principle” in Probability
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Does this probability distribution have a name?A, B, C and D, and we are trying to find the probability that exactly one event occurs.Joint probability of two random variables both following poisson distributionExpectation of of joint probability distributionApplication of the pigeon hole principleTransformation of a non one-to-one probability mass function?A question in probability and combinatoricsProblem with my intuition concerning probabilities of dependent events and conditional probabilityDuel question, why shoot when $P_1 (s)+P_2 (s)=1$High F-Statistic Value and negative Probability for Granger Causality Result, Interpretation?
$begingroup$
I am having trouble showing that for a non-negative random variable $X$, $textbfE(Xcdot I(X>frac12textbfE(X)) geq frac12textbfE(X)$. Intuitively, I see why this is true -- if we reduce the values of $X$ to zero for all values below $frac12textbfE(X)$, then the "weight" of $X$ that occurs above $textbfE(X)$ should make it so that even if all the rest of the "weight" of $X$ occurs at $0$ (which would make $textbfE(X)$ as low as possible), then this would still cause the center of mass to be at least $frac12textbfE(X)$. But I'm having trouble actually showing this principle.
Any help would be greatly appreciated.
probability
asked Mar 26 at 17:10
DJGDJG
485
$endgroup$
add a comment |
$begingroup$
I am having trouble showing that for a non-negative random variable $X$, $textbfE(Xcdot I(X>frac12textbfE(X)) geq frac12textbfE(X)$. Intuitively, I see why this is true -- if we reduce the values of $X$ to zero for all values below $frac12textbfE(X)$, then the "weight" of $X$ that occurs above $textbfE(X)$ should make it so that even if all the rest of the "weight" of $X$ occurs at $0$ (which would make $textbfE(X)$ as low as possible), then this would still cause the center of mass to be at least $frac12textbfE(X)$. But I'm having trouble actually showing this principle.
Any help would be greatly appreciated.
probability
asked Mar 26 at 17:10
DJGDJG
485
$endgroup$
add a comment |
$begingroup$
I am having trouble showing that for a non-negative random variable $X$, $textbfE(Xcdot I(X>frac12textbfE(X)) geq frac12textbfE(X)$. Intuitively, I see why this is true -- if we reduce the values of $X$ to zero for all values below $frac12textbfE(X)$, then the "weight" of $X$ that occurs above $textbfE(X)$ should make it so that even if all the rest of the "weight" of $X$ occurs at $0$ (which would make $textbfE(X)$ as low as possible), then this would still cause the center of mass to be at least $frac12textbfE(X)$. But I'm having trouble actually showing this principle.
Any help would be greatly appreciated.
probability
asked Mar 26 at 17:10
DJGDJG
485
$endgroup$
I am having trouble showing that for a non-negative random variable $X$, $textbfE(Xcdot I(X>frac12textbfE(X)) geq frac12textbfE(X)$. Intuitively, I see why this is true -- if we reduce the values of $X$ to zero for all values below $frac12textbfE(X)$, then the "weight" of $X$ that occurs above $textbfE(X)$ should make it so that even if all the rest of the "weight" of $X$ occurs at $0$ (which would make $textbfE(X)$ as low as possible), then this would still cause the center of mass to be at least $frac12textbfE(X)$. But I'm having trouble actually showing this principle.
Any help would be greatly appreciated.
probability
probability
asked Mar 26 at 17:10
DJGDJG
485
asked Mar 26 at 17:10
DJGDJG
485
asked Mar 26 at 17:10
DJGDJG
485
asked Mar 26 at 17:10
DJGDJG
485
asked Mar 26 at 17:10
DJGDJG
485
485
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Perhaps this might work for you:
$ Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ as a bounded random variable
so $textbfE(Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ taking expectations
while $textbfE(Xcdot I(Xle frac12textbfE(X)) + textbfE(Xcdot I(Xgt frac12textbfE(X)) = textbfE(X)$ by linearity of expectation
so $textbfE(Xcdot I(Xgt frac12textbfE(X)) geq frac12textbfE(X)$ by subtraction
answered Mar 26 at 17:34
HenryHenry
101k482170
$endgroup$
$begingroup$
That makes sense, thank you! Now I just have to think about how to turn my intuition into something like this.
$endgroup$
– DJG
Mar 26 at 17:47
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
$begingroup$
Perhaps this might work for you:
$ Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ as a bounded random variable
so $textbfE(Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ taking expectations
while $textbfE(Xcdot I(Xle frac12textbfE(X)) + textbfE(Xcdot I(Xgt frac12textbfE(X)) = textbfE(X)$ by linearity of expectation
so $textbfE(Xcdot I(Xgt frac12textbfE(X)) geq frac12textbfE(X)$ by subtraction
answered Mar 26 at 17:34
HenryHenry
101k482170
$endgroup$
$begingroup$
That makes sense, thank you! Now I just have to think about how to turn my intuition into something like this.
$endgroup$
– DJG
Mar 26 at 17:47
add a comment |
$begingroup$
Perhaps this might work for you:
$ Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ as a bounded random variable
so $textbfE(Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ taking expectations
while $textbfE(Xcdot I(Xle frac12textbfE(X)) + textbfE(Xcdot I(Xgt frac12textbfE(X)) = textbfE(X)$ by linearity of expectation
so $textbfE(Xcdot I(Xgt frac12textbfE(X)) geq frac12textbfE(X)$ by subtraction
answered Mar 26 at 17:34
HenryHenry
101k482170
$endgroup$
$begingroup$
That makes sense, thank you! Now I just have to think about how to turn my intuition into something like this.
$endgroup$
– DJG
Mar 26 at 17:47
add a comment |
$begingroup$
Perhaps this might work for you:
$ Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ as a bounded random variable
so $textbfE(Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ taking expectations
while $textbfE(Xcdot I(Xle frac12textbfE(X)) + textbfE(Xcdot I(Xgt frac12textbfE(X)) = textbfE(X)$ by linearity of expectation
so $textbfE(Xcdot I(Xgt frac12textbfE(X)) geq frac12textbfE(X)$ by subtraction
answered Mar 26 at 17:34
HenryHenry
101k482170
$endgroup$
Perhaps this might work for you:
$ Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ as a bounded random variable
so $textbfE(Xcdot I(Xle frac12textbfE(X)) leq frac12textbfE(X)$ taking expectations
while $textbfE(Xcdot I(Xle frac12textbfE(X)) + textbfE(Xcdot I(Xgt frac12textbfE(X)) = textbfE(X)$ by linearity of expectation
so $textbfE(Xcdot I(Xgt frac12textbfE(X)) geq frac12textbfE(X)$ by subtraction
answered Mar 26 at 17:34
HenryHenry
101k482170
answered Mar 26 at 17:34
HenryHenry
101k482170
answered Mar 26 at 17:34
HenryHenry
101k482170
answered Mar 26 at 17:34
HenryHenry
101k482170
101k482170
$begingroup$
That makes sense, thank you! Now I just have to think about how to turn my intuition into something like this.
$endgroup$
– DJG
Mar 26 at 17:47
add a comment |
$begingroup$
That makes sense, thank you! Now I just have to think about how to turn my intuition into something like this.
$endgroup$
– DJG
Mar 26 at 17:47
$begingroup$
That makes sense, thank you! Now I just have to think about how to turn my intuition into something like this.
$endgroup$
– DJG
Mar 26 at 17:47
$begingroup$
That makes sense, thank you! Now I just have to think about how to turn my intuition into something like this.
$endgroup$
– DJG
Mar 26 at 17:47
add a comment |
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