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







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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


















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    $E[X]=0$ not $0.6$. So the probablity you're calculating is $P(|X|ge.2)$
    $endgroup$
    – Anvit
    Nov 25 '18 at 7:07










  • $begingroup$
    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
















2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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




















  • $begingroup$
    $E[X]=0$ 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












1 Answer
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$begingroup$

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.






share|cite|improve this answer











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    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 '18 at 7:24

























        answered Nov 25 '18 at 7:19









        Zachary SelkZachary Selk

        592311




        592311






























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