Finding the probability of a random variable given its expectation












0












$begingroup$


I'm taking the Algorithms: Design and Analysis II class, one of the questions asks:




Suppose $ X $ is a random variable that has expected value $ 1 $.



a) What is the probability that $ X $ is $ 2 $ or larger? (Choose the
strongest statement that is guaranteed to be true.)




  • At most $ 100% $

  • At most $ 25% $

  • $ 0 $

  • At most $ 50% $


b) Does the answer change if $ X $ is always nonnegative?




This is not an active homework question. I've completed the course successfully, but I don't like loose ends, hence going back and trying to solve the problems that I couldn't answer at that time.






probability random-variables expected-value







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What have you tried? Can you eliminate some of the proposed statements?
    $endgroup$
    – lulu
    Jan 16 at 0:27










  • $begingroup$
    How is asking for the answers going back and solving the problems?
    $endgroup$
    – John Douma
    Jan 16 at 0:30










  • $begingroup$
    @lulu No, I can't.
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:20










  • $begingroup$
    @JohnDouma There's a thing called "understanding". It helps.
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:20










  • $begingroup$
    Really? Surely you can eliminate the case $P(X≥2)=0$. Just try.
    $endgroup$
    – lulu
    Jan 16 at 1:22
















0












$begingroup$


I'm taking the Algorithms: Design and Analysis II class, one of the questions asks:




Suppose $ X $ is a random variable that has expected value $ 1 $.



a) What is the probability that $ X $ is $ 2 $ or larger? (Choose the
strongest statement that is guaranteed to be true.)




  • At most $ 100% $

  • At most $ 25% $

  • $ 0 $

  • At most $ 50% $


b) Does the answer change if $ X $ is always nonnegative?




This is not an active homework question. I've completed the course successfully, but I don't like loose ends, hence going back and trying to solve the problems that I couldn't answer at that time.






probability random-variables expected-value







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What have you tried? Can you eliminate some of the proposed statements?
    $endgroup$
    – lulu
    Jan 16 at 0:27










  • $begingroup$
    How is asking for the answers going back and solving the problems?
    $endgroup$
    – John Douma
    Jan 16 at 0:30










  • $begingroup$
    @lulu No, I can't.
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:20










  • $begingroup$
    @JohnDouma There's a thing called "understanding". It helps.
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:20










  • $begingroup$
    Really? Surely you can eliminate the case $P(X≥2)=0$. Just try.
    $endgroup$
    – lulu
    Jan 16 at 1:22














0












0








0





$begingroup$


I'm taking the Algorithms: Design and Analysis II class, one of the questions asks:




Suppose $ X $ is a random variable that has expected value $ 1 $.



a) What is the probability that $ X $ is $ 2 $ or larger? (Choose the
strongest statement that is guaranteed to be true.)




  • At most $ 100% $

  • At most $ 25% $

  • $ 0 $

  • At most $ 50% $


b) Does the answer change if $ X $ is always nonnegative?




This is not an active homework question. I've completed the course successfully, but I don't like loose ends, hence going back and trying to solve the problems that I couldn't answer at that time.






probability random-variables expected-value







share|cite|improve this question









$endgroup$




I'm taking the Algorithms: Design and Analysis II class, one of the questions asks:




Suppose $ X $ is a random variable that has expected value $ 1 $.



a) What is the probability that $ X $ is $ 2 $ or larger? (Choose the
strongest statement that is guaranteed to be true.)




  • At most $ 100% $

  • At most $ 25% $

  • $ 0 $

  • At most $ 50% $


b) Does the answer change if $ X $ is always nonnegative?




This is not an active homework question. I've completed the course successfully, but I don't like loose ends, hence going back and trying to solve the problems that I couldn't answer at that time.






probability random-variables expected-value




probability random-variables expected-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 0:14









Abhijit SarkarAbhijit Sarkar

1033




1033








  • 1




    $begingroup$
    What have you tried? Can you eliminate some of the proposed statements?
    $endgroup$
    – lulu
    Jan 16 at 0:27










  • $begingroup$
    How is asking for the answers going back and solving the problems?
    $endgroup$
    – John Douma
    Jan 16 at 0:30










  • $begingroup$
    @lulu No, I can't.
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:20










  • $begingroup$
    @JohnDouma There's a thing called "understanding". It helps.
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:20










  • $begingroup$
    Really? Surely you can eliminate the case $P(X≥2)=0$. Just try.
    $endgroup$
    – lulu
    Jan 16 at 1:22














  • 1




    $begingroup$
    What have you tried? Can you eliminate some of the proposed statements?
    $endgroup$
    – lulu
    Jan 16 at 0:27










  • $begingroup$
    How is asking for the answers going back and solving the problems?
    $endgroup$
    – John Douma
    Jan 16 at 0:30










  • $begingroup$
    @lulu No, I can't.
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:20










  • $begingroup$
    @JohnDouma There's a thing called "understanding". It helps.
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:20










  • $begingroup$
    Really? Surely you can eliminate the case $P(X≥2)=0$. Just try.
    $endgroup$
    – lulu
    Jan 16 at 1:22








1




1




$begingroup$
What have you tried? Can you eliminate some of the proposed statements?
$endgroup$
– lulu
Jan 16 at 0:27




$begingroup$
What have you tried? Can you eliminate some of the proposed statements?
$endgroup$
– lulu
Jan 16 at 0:27












$begingroup$
How is asking for the answers going back and solving the problems?
$endgroup$
– John Douma
Jan 16 at 0:30




$begingroup$
How is asking for the answers going back and solving the problems?
$endgroup$
– John Douma
Jan 16 at 0:30












$begingroup$
@lulu No, I can't.
$endgroup$
– Abhijit Sarkar
Jan 16 at 1:20




$begingroup$
@lulu No, I can't.
$endgroup$
– Abhijit Sarkar
Jan 16 at 1:20












$begingroup$
@JohnDouma There's a thing called "understanding". It helps.
$endgroup$
– Abhijit Sarkar
Jan 16 at 1:20




$begingroup$
@JohnDouma There's a thing called "understanding". It helps.
$endgroup$
– Abhijit Sarkar
Jan 16 at 1:20












$begingroup$
Really? Surely you can eliminate the case $P(X≥2)=0$. Just try.
$endgroup$
– lulu
Jan 16 at 1:22




$begingroup$
Really? Surely you can eliminate the case $P(X≥2)=0$. Just try.
$endgroup$
– lulu
Jan 16 at 1:22










1 Answer
1






active

oldest

votes


















2












$begingroup$

For part $b$, using Markov's Inequality we have
$$P(Xgeq 2)leq frac{E[X]}{2}=frac{1}{2}$$



For part $a$, consider the discrete random variable defined by
$$P(X=2)=.9$$
$$P(X=-8)=.1$$
Then $E(X)=1$ yet $P(Xgeq 2)=.9$. This counterexample eliminates the second, third, and fourth options. So it must be the (trivial) first option: "at most $100%"$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there a way to prove part a formally, without using a counterexample? The thing with counterexamples is, they are only as good as the person providing them. Who's to say there isn't a counterexample that invalidates option 1, and none of the answers are correct? (in other words, we seem to be cheating a little knowing the fact that one of the four options must be correct; how'd you proceed if only the question was given, not the options).
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:37










  • $begingroup$
    I'm not sure what you mean, a counterexample is perfectly valid here. And there can be no counterexample for option $1$, since obviously no probability can be higher than 100%
    $endgroup$
    – pwerth
    Jan 16 at 2:04










  • $begingroup$
    A counterexample is valid, but not proven to be exhaustive. What if a tighter bound, say 90% exists?
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:52










  • $begingroup$
    Your question says "Choose the strongest statement that is guaranteed to be true." My counterexample shows that "0", "at most 25%" and "at most 50%" are not guaranteed to be true. Since "at most 100%" is always true, it is the only possible answer for part (a).
    $endgroup$
    – pwerth
    Jan 16 at 2:56












  • $begingroup$
    Agreed, which is why I accepted and upvoted your answer. However, I'm curious if a more formal treatment is possible to, "Prove the tightest possible bound"
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:57














Your Answer





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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

For part $b$, using Markov's Inequality we have
$$P(Xgeq 2)leq frac{E[X]}{2}=frac{1}{2}$$



For part $a$, consider the discrete random variable defined by
$$P(X=2)=.9$$
$$P(X=-8)=.1$$
Then $E(X)=1$ yet $P(Xgeq 2)=.9$. This counterexample eliminates the second, third, and fourth options. So it must be the (trivial) first option: "at most $100%"$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there a way to prove part a formally, without using a counterexample? The thing with counterexamples is, they are only as good as the person providing them. Who's to say there isn't a counterexample that invalidates option 1, and none of the answers are correct? (in other words, we seem to be cheating a little knowing the fact that one of the four options must be correct; how'd you proceed if only the question was given, not the options).
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:37










  • $begingroup$
    I'm not sure what you mean, a counterexample is perfectly valid here. And there can be no counterexample for option $1$, since obviously no probability can be higher than 100%
    $endgroup$
    – pwerth
    Jan 16 at 2:04










  • $begingroup$
    A counterexample is valid, but not proven to be exhaustive. What if a tighter bound, say 90% exists?
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:52










  • $begingroup$
    Your question says "Choose the strongest statement that is guaranteed to be true." My counterexample shows that "0", "at most 25%" and "at most 50%" are not guaranteed to be true. Since "at most 100%" is always true, it is the only possible answer for part (a).
    $endgroup$
    – pwerth
    Jan 16 at 2:56












  • $begingroup$
    Agreed, which is why I accepted and upvoted your answer. However, I'm curious if a more formal treatment is possible to, "Prove the tightest possible bound"
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:57


















2












$begingroup$

For part $b$, using Markov's Inequality we have
$$P(Xgeq 2)leq frac{E[X]}{2}=frac{1}{2}$$



For part $a$, consider the discrete random variable defined by
$$P(X=2)=.9$$
$$P(X=-8)=.1$$
Then $E(X)=1$ yet $P(Xgeq 2)=.9$. This counterexample eliminates the second, third, and fourth options. So it must be the (trivial) first option: "at most $100%"$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there a way to prove part a formally, without using a counterexample? The thing with counterexamples is, they are only as good as the person providing them. Who's to say there isn't a counterexample that invalidates option 1, and none of the answers are correct? (in other words, we seem to be cheating a little knowing the fact that one of the four options must be correct; how'd you proceed if only the question was given, not the options).
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:37










  • $begingroup$
    I'm not sure what you mean, a counterexample is perfectly valid here. And there can be no counterexample for option $1$, since obviously no probability can be higher than 100%
    $endgroup$
    – pwerth
    Jan 16 at 2:04










  • $begingroup$
    A counterexample is valid, but not proven to be exhaustive. What if a tighter bound, say 90% exists?
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:52










  • $begingroup$
    Your question says "Choose the strongest statement that is guaranteed to be true." My counterexample shows that "0", "at most 25%" and "at most 50%" are not guaranteed to be true. Since "at most 100%" is always true, it is the only possible answer for part (a).
    $endgroup$
    – pwerth
    Jan 16 at 2:56












  • $begingroup$
    Agreed, which is why I accepted and upvoted your answer. However, I'm curious if a more formal treatment is possible to, "Prove the tightest possible bound"
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:57
















2












2








2





$begingroup$

For part $b$, using Markov's Inequality we have
$$P(Xgeq 2)leq frac{E[X]}{2}=frac{1}{2}$$



For part $a$, consider the discrete random variable defined by
$$P(X=2)=.9$$
$$P(X=-8)=.1$$
Then $E(X)=1$ yet $P(Xgeq 2)=.9$. This counterexample eliminates the second, third, and fourth options. So it must be the (trivial) first option: "at most $100%"$






share|cite|improve this answer











$endgroup$



For part $b$, using Markov's Inequality we have
$$P(Xgeq 2)leq frac{E[X]}{2}=frac{1}{2}$$



For part $a$, consider the discrete random variable defined by
$$P(X=2)=.9$$
$$P(X=-8)=.1$$
Then $E(X)=1$ yet $P(Xgeq 2)=.9$. This counterexample eliminates the second, third, and fourth options. So it must be the (trivial) first option: "at most $100%"$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 0:31

























answered Jan 16 at 0:27









pwerthpwerth

3,300417




3,300417












  • $begingroup$
    Is there a way to prove part a formally, without using a counterexample? The thing with counterexamples is, they are only as good as the person providing them. Who's to say there isn't a counterexample that invalidates option 1, and none of the answers are correct? (in other words, we seem to be cheating a little knowing the fact that one of the four options must be correct; how'd you proceed if only the question was given, not the options).
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:37










  • $begingroup$
    I'm not sure what you mean, a counterexample is perfectly valid here. And there can be no counterexample for option $1$, since obviously no probability can be higher than 100%
    $endgroup$
    – pwerth
    Jan 16 at 2:04










  • $begingroup$
    A counterexample is valid, but not proven to be exhaustive. What if a tighter bound, say 90% exists?
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:52










  • $begingroup$
    Your question says "Choose the strongest statement that is guaranteed to be true." My counterexample shows that "0", "at most 25%" and "at most 50%" are not guaranteed to be true. Since "at most 100%" is always true, it is the only possible answer for part (a).
    $endgroup$
    – pwerth
    Jan 16 at 2:56












  • $begingroup$
    Agreed, which is why I accepted and upvoted your answer. However, I'm curious if a more formal treatment is possible to, "Prove the tightest possible bound"
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:57




















  • $begingroup$
    Is there a way to prove part a formally, without using a counterexample? The thing with counterexamples is, they are only as good as the person providing them. Who's to say there isn't a counterexample that invalidates option 1, and none of the answers are correct? (in other words, we seem to be cheating a little knowing the fact that one of the four options must be correct; how'd you proceed if only the question was given, not the options).
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 1:37










  • $begingroup$
    I'm not sure what you mean, a counterexample is perfectly valid here. And there can be no counterexample for option $1$, since obviously no probability can be higher than 100%
    $endgroup$
    – pwerth
    Jan 16 at 2:04










  • $begingroup$
    A counterexample is valid, but not proven to be exhaustive. What if a tighter bound, say 90% exists?
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:52










  • $begingroup$
    Your question says "Choose the strongest statement that is guaranteed to be true." My counterexample shows that "0", "at most 25%" and "at most 50%" are not guaranteed to be true. Since "at most 100%" is always true, it is the only possible answer for part (a).
    $endgroup$
    – pwerth
    Jan 16 at 2:56












  • $begingroup$
    Agreed, which is why I accepted and upvoted your answer. However, I'm curious if a more formal treatment is possible to, "Prove the tightest possible bound"
    $endgroup$
    – Abhijit Sarkar
    Jan 16 at 2:57


















$begingroup$
Is there a way to prove part a formally, without using a counterexample? The thing with counterexamples is, they are only as good as the person providing them. Who's to say there isn't a counterexample that invalidates option 1, and none of the answers are correct? (in other words, we seem to be cheating a little knowing the fact that one of the four options must be correct; how'd you proceed if only the question was given, not the options).
$endgroup$
– Abhijit Sarkar
Jan 16 at 1:37




$begingroup$
Is there a way to prove part a formally, without using a counterexample? The thing with counterexamples is, they are only as good as the person providing them. Who's to say there isn't a counterexample that invalidates option 1, and none of the answers are correct? (in other words, we seem to be cheating a little knowing the fact that one of the four options must be correct; how'd you proceed if only the question was given, not the options).
$endgroup$
– Abhijit Sarkar
Jan 16 at 1:37












$begingroup$
I'm not sure what you mean, a counterexample is perfectly valid here. And there can be no counterexample for option $1$, since obviously no probability can be higher than 100%
$endgroup$
– pwerth
Jan 16 at 2:04




$begingroup$
I'm not sure what you mean, a counterexample is perfectly valid here. And there can be no counterexample for option $1$, since obviously no probability can be higher than 100%
$endgroup$
– pwerth
Jan 16 at 2:04












$begingroup$
A counterexample is valid, but not proven to be exhaustive. What if a tighter bound, say 90% exists?
$endgroup$
– Abhijit Sarkar
Jan 16 at 2:52




$begingroup$
A counterexample is valid, but not proven to be exhaustive. What if a tighter bound, say 90% exists?
$endgroup$
– Abhijit Sarkar
Jan 16 at 2:52












$begingroup$
Your question says "Choose the strongest statement that is guaranteed to be true." My counterexample shows that "0", "at most 25%" and "at most 50%" are not guaranteed to be true. Since "at most 100%" is always true, it is the only possible answer for part (a).
$endgroup$
– pwerth
Jan 16 at 2:56






$begingroup$
Your question says "Choose the strongest statement that is guaranteed to be true." My counterexample shows that "0", "at most 25%" and "at most 50%" are not guaranteed to be true. Since "at most 100%" is always true, it is the only possible answer for part (a).
$endgroup$
– pwerth
Jan 16 at 2:56














$begingroup$
Agreed, which is why I accepted and upvoted your answer. However, I'm curious if a more formal treatment is possible to, "Prove the tightest possible bound"
$endgroup$
– Abhijit Sarkar
Jan 16 at 2:57






$begingroup$
Agreed, which is why I accepted and upvoted your answer. However, I'm curious if a more formal treatment is possible to, "Prove the tightest possible bound"
$endgroup$
– Abhijit Sarkar
Jan 16 at 2:57




















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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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