Markov inequality for random variables with negative values.

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I'm given the maximum value of a random variable $X$ (for example $50$) and its mean, $mathbb E(X)=20$. How do I find the upper bound to $P(Xle -10)$?

edited Nov 14 at 8:32

Jimmy R.

32.9k42157

asked Nov 14 at 6:21

puffles

669

1

I think that you need to make the transformation $Y = X + 11$ and work from there.
– Ekesh
Nov 14 at 8:37

add a comment |

up vote
2
down vote

favorite

I'm given the maximum value of a random variable $X$ (for example $50$) and its mean, $mathbb E(X)=20$. How do I find the upper bound to $P(Xle -10)$?

edited Nov 14 at 8:32

Jimmy R.

32.9k42157

asked Nov 14 at 6:21

puffles

669

1

I think that you need to make the transformation $Y = X + 11$ and work from there.
– Ekesh
Nov 14 at 8:37

add a comment |

up vote
2
down vote

favorite

I'm given the maximum value of a random variable $X$ (for example $50$) and its mean, $mathbb E(X)=20$. How do I find the upper bound to $P(Xle -10)$?

edited Nov 14 at 8:32

Jimmy R.

32.9k42157

asked Nov 14 at 6:21

puffles

669

I'm given the maximum value of a random variable $X$ (for example $50$) and its mean, $mathbb E(X)=20$. How do I find the upper bound to $P(Xle -10)$?

random-variables

edited Nov 14 at 8:32

Jimmy R.

32.9k42157

asked Nov 14 at 6:21

puffles

669

edited Nov 14 at 8:32

Jimmy R.

32.9k42157

asked Nov 14 at 6:21

puffles

669

edited Nov 14 at 8:32

Jimmy R.

32.9k42157

edited Nov 14 at 8:32

Jimmy R.

32.9k42157

edited Nov 14 at 8:32

Jimmy R.

32.9k42157

asked Nov 14 at 6:21

puffles

669

asked Nov 14 at 6:21

puffles

669

asked Nov 14 at 6:21

puffles

669

1

I think that you need to make the transformation $Y = X + 11$ and work from there.
– Ekesh
Nov 14 at 8:37

add a comment |

1

I think that you need to make the transformation $Y = X + 11$ and work from there.
– Ekesh
Nov 14 at 8:37

I think that you need to make the transformation $Y = X + 11$ and work from there.
– Ekesh
Nov 14 at 8:37

add a comment |

2 Answers
2

active

oldest

votes

up vote
0
down vote

accepted

Define new RV: S = 50 - R

E(S) = 30

P(R<=-10) = P(S>=60)

P(S>=60) = 30/60
= 1/2

You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3

It should be 1/2 not 1/3.

answered Nov 17 at 20:17

helloworld

477

How is the expectation 30. Can you please explain?
– puffles
Nov 17 at 20:53

S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
– helloworld
Nov 17 at 21:24

add a comment |

up vote
1
down vote

Hint:

$50-X$ is a nonnegative random variable since $50$ is an upperbound.

Express your inequality in the form of $Pr(50-X ge c)$.

answered Nov 14 at 8:38

Siong Thye Goh

94k1462114

So it should be like this : P(50-X >= -10) ?
– puffles
Nov 14 at 11:23

$P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
– Siong Thye Goh
Nov 14 at 11:24

Makes sense. Thanks a bunch!
– puffles
Nov 14 at 11:25

I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
– puffles
Nov 15 at 15:26

1

yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
– Siong Thye Goh
Nov 15 at 15:29

|
show 1 more comment

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
0
down vote

accepted

Define new RV: S = 50 - R

E(S) = 30

P(R<=-10) = P(S>=60)

P(S>=60) = 30/60
= 1/2

You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3

It should be 1/2 not 1/3.

answered Nov 17 at 20:17

helloworld

477

How is the expectation 30. Can you please explain?
– puffles
Nov 17 at 20:53

S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
– helloworld
Nov 17 at 21:24

add a comment |

up vote
0
down vote

accepted

Define new RV: S = 50 - R

E(S) = 30

P(R<=-10) = P(S>=60)

P(S>=60) = 30/60
= 1/2

You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3

It should be 1/2 not 1/3.

answered Nov 17 at 20:17

helloworld

477

How is the expectation 30. Can you please explain?
– puffles
Nov 17 at 20:53

S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
– helloworld
Nov 17 at 21:24

add a comment |

up vote
0
down vote

accepted

Define new RV: S = 50 - R

E(S) = 30

P(R<=-10) = P(S>=60)

P(S>=60) = 30/60
= 1/2

You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3

It should be 1/2 not 1/3.

answered Nov 17 at 20:17

helloworld

477

Define new RV: S = 50 - R

E(S) = 30

P(R<=-10) = P(S>=60)

P(S>=60) = 30/60
= 1/2

You commented: I needed to clarify one last thing. P(50-X >= 60) <= 1/3

It should be 1/2 not 1/3.

answered Nov 17 at 20:17

helloworld

477

answered Nov 17 at 20:17

helloworld

477

answered Nov 17 at 20:17

helloworld

477

answered Nov 17 at 20:17

helloworld

477

How is the expectation 30. Can you please explain?
– puffles
Nov 17 at 20:53

S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
– helloworld
Nov 17 at 21:24

add a comment |

How is the expectation 30. Can you please explain?
– puffles
Nov 17 at 20:53

S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
– helloworld
Nov 17 at 21:24

How is the expectation 30. Can you please explain?
– puffles
Nov 17 at 20:53

S = 50 -R so E(S) = 50 -E(R) . Since E(R) = 20. We get E(S) = 30
– helloworld
Nov 17 at 21:24

add a comment |

up vote
1
down vote

Hint:

$50-X$ is a nonnegative random variable since $50$ is an upperbound.

Express your inequality in the form of $Pr(50-X ge c)$.

answered Nov 14 at 8:38

Siong Thye Goh

94k1462114

So it should be like this : P(50-X >= -10) ?
– puffles
Nov 14 at 11:23

$P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
– Siong Thye Goh
Nov 14 at 11:24

Makes sense. Thanks a bunch!
– puffles
Nov 14 at 11:25

I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
– puffles
Nov 15 at 15:26

1

yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
– Siong Thye Goh
Nov 15 at 15:29

|
show 1 more comment

up vote
1
down vote

Hint:

$50-X$ is a nonnegative random variable since $50$ is an upperbound.

Express your inequality in the form of $Pr(50-X ge c)$.

answered Nov 14 at 8:38

Siong Thye Goh

94k1462114

So it should be like this : P(50-X >= -10) ?
– puffles
Nov 14 at 11:23

$P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
– Siong Thye Goh
Nov 14 at 11:24

Makes sense. Thanks a bunch!
– puffles
Nov 14 at 11:25

I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
– puffles
Nov 15 at 15:26

1

yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
– Siong Thye Goh
Nov 15 at 15:29

|
show 1 more comment

up vote
1
down vote

Hint:

$50-X$ is a nonnegative random variable since $50$ is an upperbound.

Express your inequality in the form of $Pr(50-X ge c)$.

answered Nov 14 at 8:38

Siong Thye Goh

94k1462114

Hint:

$50-X$ is a nonnegative random variable since $50$ is an upperbound.

Express your inequality in the form of $Pr(50-X ge c)$.

answered Nov 14 at 8:38

Siong Thye Goh

94k1462114

answered Nov 14 at 8:38

Siong Thye Goh

94k1462114

answered Nov 14 at 8:38

Siong Thye Goh

94k1462114

answered Nov 14 at 8:38

Siong Thye Goh

94k1462114

So it should be like this : P(50-X >= -10) ?
– puffles
Nov 14 at 11:23

$P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
– Siong Thye Goh
Nov 14 at 11:24

Makes sense. Thanks a bunch!
– puffles
Nov 14 at 11:25

I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
– puffles
Nov 15 at 15:26

1

yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
– Siong Thye Goh
Nov 15 at 15:29

|
show 1 more comment

So it should be like this : P(50-X >= -10) ?
– puffles
Nov 14 at 11:23

$P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
– Siong Thye Goh
Nov 14 at 11:24

Makes sense. Thanks a bunch!
– puffles
Nov 14 at 11:25

I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
– puffles
Nov 15 at 15:26

1

yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
– Siong Thye Goh
Nov 15 at 15:29

So it should be like this : P(50-X >= -10) ?
– puffles
Nov 14 at 11:23

$P(X le -10) = P(-X ge 10) = P(50-X ge 60)$, now apply Markov on $50-X$.
– Siong Thye Goh
Nov 14 at 11:24

Makes sense. Thanks a bunch!
– puffles
Nov 14 at 11:25

I needed to clarify one last thing. P(50-X >= 60) <= 1/3. Is this correct considering the typical formula of Markov inequality P(X >= a)? What I am trying to ask is that if a constant is added or subtracted in an interval such as in the above case, would it affect the final probability.
– puffles
Nov 15 at 15:26

yes, it is correct. if it makes you more comfortable, let $Y=50-X$ and check that $Y$ is nonnegative. If you perform some operations and prove that the two conditions are equivalent, then the probability stays the same.
– Siong Thye Goh
Nov 15 at 15:29

|
show 1 more comment

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