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
asked Jan 16 at 0:14
Abhijit SarkarAbhijit Sarkar
1033
$endgroup$
add a comment |
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
asked Jan 16 at 0:14
Abhijit SarkarAbhijit Sarkar
1033
$endgroup$
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
add a comment |
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
asked Jan 16 at 0:14
Abhijit SarkarAbhijit Sarkar
1033
$endgroup$
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
asked Jan 16 at 0:14
Abhijit SarkarAbhijit Sarkar
1033
asked Jan 16 at 0:14
Abhijit SarkarAbhijit Sarkar
1033
asked Jan 16 at 0:14
Abhijit SarkarAbhijit Sarkar
1033
asked Jan 16 at 0:14
Abhijit SarkarAbhijit Sarkar
1033
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
add a comment |
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
add a comment |
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 $100%"$
answered Jan 16 at 0:27
pwerthpwerth
3,300417
$endgroup$
$begingroup$
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
|
show 2 more comments
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1 Answer
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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 $100%"$
answered Jan 16 at 0:27
pwerthpwerth
3,300417
$endgroup$
$begingroup$
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
|
show 2 more comments
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 $100%"$
answered Jan 16 at 0:27
pwerthpwerth
3,300417
$endgroup$
$begingroup$
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
|
show 2 more comments
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 $100%"$
answered Jan 16 at 0:27
pwerthpwerth
3,300417
$endgroup$
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 $100%"$
answered Jan 16 at 0:27
pwerthpwerth
3,300417
edited Jan 16 at 0:31
answered Jan 16 at 0:27
pwerthpwerth
3,300417
answered Jan 16 at 0:27
pwerthpwerth
3,300417
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
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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