Deriving the probability on the Probability of an Event - Chebyshev's Theorem
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Question: For the random variable X, where E(X) = 0 and Var(X) = 1, derive the upper bound on the probability of the event ${|X-.6| > .2}$.
My attempt: Using Chebyshev's Inequality- Prob($|X-E(X)| > epsilon) leq frac{Var(X)}{epsilon^2} = frac{1}{(.2)^2} = 25.$
A) With no distributional assumption is this correct? I am thinking that because the probability is greater than 1 then this is a trivial inequality. I have seen alternate definitions of Chebyshev's inequality where $epsilon$ is $k sigma$ and if k is greater than or equal to 1 then this happens. However, 25 seems like an unusual number to get. Am I missing something?
B) As a follow up and for further understanding, how would the probability change if X was normally distributed with mean = 0 and variance = 1? Using the probability normal approximation integral I get the probability to be 3.96953... Although 25 and 3.96 are very different, is accuracy even a question since both probabilities are greater than 1?
probability statistics
asked Nov 25 '18 at 6:34
JoeJoe
132
$endgroup$
add a comment |
$begingroup$
Question: For the random variable X, where E(X) = 0 and Var(X) = 1, derive the upper bound on the probability of the event ${|X-.6| > .2}$.
My attempt: Using Chebyshev's Inequality- Prob($|X-E(X)| > epsilon) leq frac{Var(X)}{epsilon^2} = frac{1}{(.2)^2} = 25.$
A) With no distributional assumption is this correct? I am thinking that because the probability is greater than 1 then this is a trivial inequality. I have seen alternate definitions of Chebyshev's inequality where $epsilon$ is $k sigma$ and if k is greater than or equal to 1 then this happens. However, 25 seems like an unusual number to get. Am I missing something?
B) As a follow up and for further understanding, how would the probability change if X was normally distributed with mean = 0 and variance = 1? Using the probability normal approximation integral I get the probability to be 3.96953... Although 25 and 3.96 are very different, is accuracy even a question since both probabilities are greater than 1?
probability statistics
asked Nov 25 '18 at 6:34
JoeJoe
132
$endgroup$
$begingroup$
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
$endgroup$
– Anvit
Nov 25 '18 at 7:07
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Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
$endgroup$
– Anvit
Nov 25 '18 at 7:11
add a comment |
$begingroup$
Question: For the random variable X, where E(X) = 0 and Var(X) = 1, derive the upper bound on the probability of the event ${|X-.6| > .2}$.
My attempt: Using Chebyshev's Inequality- Prob($|X-E(X)| > epsilon) leq frac{Var(X)}{epsilon^2} = frac{1}{(.2)^2} = 25.$
A) With no distributional assumption is this correct? I am thinking that because the probability is greater than 1 then this is a trivial inequality. I have seen alternate definitions of Chebyshev's inequality where $epsilon$ is $k sigma$ and if k is greater than or equal to 1 then this happens. However, 25 seems like an unusual number to get. Am I missing something?
B) As a follow up and for further understanding, how would the probability change if X was normally distributed with mean = 0 and variance = 1? Using the probability normal approximation integral I get the probability to be 3.96953... Although 25 and 3.96 are very different, is accuracy even a question since both probabilities are greater than 1?
probability statistics
asked Nov 25 '18 at 6:34
JoeJoe
132
$endgroup$
Question: For the random variable X, where E(X) = 0 and Var(X) = 1, derive the upper bound on the probability of the event ${|X-.6| > .2}$.
My attempt: Using Chebyshev's Inequality- Prob($|X-E(X)| > epsilon) leq frac{Var(X)}{epsilon^2} = frac{1}{(.2)^2} = 25.$
A) With no distributional assumption is this correct? I am thinking that because the probability is greater than 1 then this is a trivial inequality. I have seen alternate definitions of Chebyshev's inequality where $epsilon$ is $k sigma$ and if k is greater than or equal to 1 then this happens. However, 25 seems like an unusual number to get. Am I missing something?
B) As a follow up and for further understanding, how would the probability change if X was normally distributed with mean = 0 and variance = 1? Using the probability normal approximation integral I get the probability to be 3.96953... Although 25 and 3.96 are very different, is accuracy even a question since both probabilities are greater than 1?
probability statistics
probability statistics
asked Nov 25 '18 at 6:34
JoeJoe
132
asked Nov 25 '18 at 6:34
JoeJoe
132
asked Nov 25 '18 at 6:34
JoeJoe
132
asked Nov 25 '18 at 6:34
JoeJoe
132
asked Nov 25 '18 at 6:34
JoeJoe
132
132
$begingroup$
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
$endgroup$
– Anvit
Nov 25 '18 at 7:07
$begingroup$
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
$endgroup$
– Anvit
Nov 25 '18 at 7:11
add a comment |
$begingroup$
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
$endgroup$
– Anvit
Nov 25 '18 at 7:07
$begingroup$
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
$endgroup$
– Anvit
Nov 25 '18 at 7:11
$begingroup$
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
$endgroup$
– Anvit
Nov 25 '18 at 7:07
$begingroup$
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
$endgroup$
– Anvit
Nov 25 '18 at 7:07
$begingroup$
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
$endgroup$
– Anvit
Nov 25 '18 at 7:11
$begingroup$
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
$endgroup$
– Anvit
Nov 25 '18 at 7:11
add a comment |
1 Answer
1
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votes
$begingroup$
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
answered Nov 25 '18 at 7:19
Zachary SelkZachary Selk
592311
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add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
answered Nov 25 '18 at 7:19
Zachary SelkZachary Selk
592311
$endgroup$
add a comment |
$begingroup$
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
answered Nov 25 '18 at 7:19
Zachary SelkZachary Selk
592311
$endgroup$
add a comment |
$begingroup$
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
answered Nov 25 '18 at 7:19
Zachary SelkZachary Selk
592311
$endgroup$
You misused Chebyshev's inequality. You computed $P(|X|> .2)$ not $P(|X-.6|>.2)$. Using Markov's inequality (i.e. Chebyshev's) we get $$P(|X-.6|>.2)=P((X-.6)^2>.04)lefrac{E(X-.6)^2}{.04}=frac{1-0+.36}{.04}$$
Still, this is greater than $1$. This is to be expected. Consider $X$ to be a coinflip with $50/50$ shot of $pm 1$. Then $E(X)=0, operatorname{var}(X)=1$. And for all $varepsilon<1$ we have $$P(|X|>varepsilon)=1$$
You can modify this for the case of $|X-.6|$.
So the best bound we can get is $1$ without further information.
answered Nov 25 '18 at 7:19
Zachary SelkZachary Selk
592311
edited Nov 25 '18 at 7:24
answered Nov 25 '18 at 7:19
Zachary SelkZachary Selk
592311
answered Nov 25 '18 at 7:19
Zachary SelkZachary Selk
592311
answered Nov 25 '18 at 7:19
Zachary SelkZachary Selk
592311
592311
add a comment |
add a comment |
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$begingroup$
$E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
$endgroup$
– Anvit
Nov 25 '18 at 7:07
$begingroup$
Also, assuming normal distribution, $P(|X-.6|>.2)=P(X>0.8 or X<0.4)$ which is equal to approx $0.867$
$endgroup$
– Anvit
Nov 25 '18 at 7:11