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?
probability-theory probability-distributions
asked Nov 20 at 15:39
henceproved
1358
add a comment |
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
asked Nov 20 at 15:39
henceproved
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
add a comment |
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
asked Nov 20 at 15:39
henceproved
1358
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
probability-theory probability-distributions
asked Nov 20 at 15:39
henceproved
1358
asked Nov 20 at 15:39
henceproved
1358
asked Nov 20 at 15:39
henceproved
1358
asked Nov 20 at 15:39
henceproved
1358
asked Nov 20 at 15:39
henceproved
1358
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
add a comment |
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
add a comment |
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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