Non-unique random variable and Chebychev inequality











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Consider the random variables $X_1$ and $X_2$, both describing the experiment of tossing an unbaised coin as follows:



For $X_1$,



$X_1 = -1$ means tails and $X_1 = 1$ means heads. Hence $$E(X_1) = 0.5 times (-1) + 0.5 times (1) = 0.$$



Now I want to calculate $P(X_1 geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e.
$$ P(X_1 geq 1) leq E(X_1)/1 = 0.$$ Since Probability is always positive, we have $P(X_1 geq 1) = 0$ but the direct calculation shows $P(X_1 geq 1) = P(X_1 = 1) = 1/2$ - a contradiction to Chebychev inequality.



For $X_2$,



$X_2 = 0$ means tails and $X_2 = 1$ means heads. Hence $$E(X_2) = 0.5 times (0) + 0.5 times (1) = 1/2.$$



Now I want to calculate $P(X_2 geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e.
$$ P(X_2 geq 1) leq E(X_2)/1 = (1/2)/1 = 1/2.$$ Also a direct calculation shows $P(X_2 geq 1) = P(X_2 = 1) = 1/2$ - a verification to Chebychev inequality.



So my question is, does the chebyshev inequality is only true for 1 among many random variable available to model a given expirement. If yes how do I recognise which one is it?










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  • actually I just realized my mistake. So should I delete this or keep it for future reference for other visitors.
    – henceproved
    Nov 20 at 15:43












  • 1. You are using Markov, not Chebyshev. 2. Markov only applies to non-negative r.v.'s.
    – Clement C.
    Nov 20 at 15:45















up vote
0
down vote

favorite












Consider the random variables $X_1$ and $X_2$, both describing the experiment of tossing an unbaised coin as follows:



For $X_1$,



$X_1 = -1$ means tails and $X_1 = 1$ means heads. Hence $$E(X_1) = 0.5 times (-1) + 0.5 times (1) = 0.$$



Now I want to calculate $P(X_1 geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e.
$$ P(X_1 geq 1) leq E(X_1)/1 = 0.$$ Since Probability is always positive, we have $P(X_1 geq 1) = 0$ but the direct calculation shows $P(X_1 geq 1) = P(X_1 = 1) = 1/2$ - a contradiction to Chebychev inequality.



For $X_2$,



$X_2 = 0$ means tails and $X_2 = 1$ means heads. Hence $$E(X_2) = 0.5 times (0) + 0.5 times (1) = 1/2.$$



Now I want to calculate $P(X_2 geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e.
$$ P(X_2 geq 1) leq E(X_2)/1 = (1/2)/1 = 1/2.$$ Also a direct calculation shows $P(X_2 geq 1) = P(X_2 = 1) = 1/2$ - a verification to Chebychev inequality.



So my question is, does the chebyshev inequality is only true for 1 among many random variable available to model a given expirement. If yes how do I recognise which one is it?










share|cite|improve this question






















  • actually I just realized my mistake. So should I delete this or keep it for future reference for other visitors.
    – henceproved
    Nov 20 at 15:43












  • 1. You are using Markov, not Chebyshev. 2. Markov only applies to non-negative r.v.'s.
    – Clement C.
    Nov 20 at 15:45













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Consider the random variables $X_1$ and $X_2$, both describing the experiment of tossing an unbaised coin as follows:



For $X_1$,



$X_1 = -1$ means tails and $X_1 = 1$ means heads. Hence $$E(X_1) = 0.5 times (-1) + 0.5 times (1) = 0.$$



Now I want to calculate $P(X_1 geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e.
$$ P(X_1 geq 1) leq E(X_1)/1 = 0.$$ Since Probability is always positive, we have $P(X_1 geq 1) = 0$ but the direct calculation shows $P(X_1 geq 1) = P(X_1 = 1) = 1/2$ - a contradiction to Chebychev inequality.



For $X_2$,



$X_2 = 0$ means tails and $X_2 = 1$ means heads. Hence $$E(X_2) = 0.5 times (0) + 0.5 times (1) = 1/2.$$



Now I want to calculate $P(X_2 geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e.
$$ P(X_2 geq 1) leq E(X_2)/1 = (1/2)/1 = 1/2.$$ Also a direct calculation shows $P(X_2 geq 1) = P(X_2 = 1) = 1/2$ - a verification to Chebychev inequality.



So my question is, does the chebyshev inequality is only true for 1 among many random variable available to model a given expirement. If yes how do I recognise which one is it?










share|cite|improve this question













Consider the random variables $X_1$ and $X_2$, both describing the experiment of tossing an unbaised coin as follows:



For $X_1$,



$X_1 = -1$ means tails and $X_1 = 1$ means heads. Hence $$E(X_1) = 0.5 times (-1) + 0.5 times (1) = 0.$$



Now I want to calculate $P(X_1 geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e.
$$ P(X_1 geq 1) leq E(X_1)/1 = 0.$$ Since Probability is always positive, we have $P(X_1 geq 1) = 0$ but the direct calculation shows $P(X_1 geq 1) = P(X_1 = 1) = 1/2$ - a contradiction to Chebychev inequality.



For $X_2$,



$X_2 = 0$ means tails and $X_2 = 1$ means heads. Hence $$E(X_2) = 0.5 times (0) + 0.5 times (1) = 1/2.$$



Now I want to calculate $P(X_2 geq 1)$ (i.e. probability of heads) through Chebychev inequality i.e.
$$ P(X_2 geq 1) leq E(X_2)/1 = (1/2)/1 = 1/2.$$ Also a direct calculation shows $P(X_2 geq 1) = P(X_2 = 1) = 1/2$ - a verification to Chebychev inequality.



So my question is, does the chebyshev inequality is only true for 1 among many random variable available to model a given expirement. If yes how do I recognise which one is it?







probability-theory probability-distributions






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asked Nov 20 at 15:39









henceproved

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1358












  • actually I just realized my mistake. So should I delete this or keep it for future reference for other visitors.
    – henceproved
    Nov 20 at 15:43












  • 1. You are using Markov, not Chebyshev. 2. Markov only applies to non-negative r.v.'s.
    – Clement C.
    Nov 20 at 15:45


















  • actually I just realized my mistake. So should I delete this or keep it for future reference for other visitors.
    – henceproved
    Nov 20 at 15:43












  • 1. You are using Markov, not Chebyshev. 2. Markov only applies to non-negative r.v.'s.
    – Clement C.
    Nov 20 at 15:45
















actually I just realized my mistake. So should I delete this or keep it for future reference for other visitors.
– henceproved
Nov 20 at 15:43






actually I just realized my mistake. So should I delete this or keep it for future reference for other visitors.
– henceproved
Nov 20 at 15:43














1. You are using Markov, not Chebyshev. 2. Markov only applies to non-negative r.v.'s.
– Clement C.
Nov 20 at 15:45




1. You are using Markov, not Chebyshev. 2. Markov only applies to non-negative r.v.'s.
– Clement C.
Nov 20 at 15:45















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