Markov’s inequality and Poisson distribution [closed]
Let X be a random variable having the Poisson distribution with parameter 1. What does Markov’s inequality (p.72) imply about the probability P{X ≥ 2}?
probability poisson-distribution
edited Nov 29 '18 at 9:48
GNUSupporter 8964民主女神 地下教會
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asked Nov 29 '18 at 9:45
Yun Chien YenYun Chien Yen
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closed as off-topic by Kavi Rama Murthy, caverac, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Rebellos Nov 29 '18 at 12:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let X be a random variable having the Poisson distribution with parameter 1. What does Markov’s inequality (p.72) imply about the probability P{X ≥ 2}?
probability poisson-distribution
edited Nov 29 '18 at 9:48
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 '18 at 9:45
Yun Chien YenYun Chien Yen
31
closed as off-topic by Kavi Rama Murthy, caverac, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Rebellos Nov 29 '18 at 12:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, caverac, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
Markov's inequality applies to all r.v., so it utilizes little characteristics of a specific probability distribution. As a result, the upper bound given by Markov/Chebylshev inequality is often too weak.
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:50
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 29 '18 at 9:50
add a comment |
Let X be a random variable having the Poisson distribution with parameter 1. What does Markov’s inequality (p.72) imply about the probability P{X ≥ 2}?
probability poisson-distribution
edited Nov 29 '18 at 9:48
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 '18 at 9:45
Yun Chien YenYun Chien Yen
31
Let X be a random variable having the Poisson distribution with parameter 1. What does Markov’s inequality (p.72) imply about the probability P{X ≥ 2}?
probability poisson-distribution
probability poisson-distribution
edited Nov 29 '18 at 9:48
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 '18 at 9:45
Yun Chien YenYun Chien Yen
31
edited Nov 29 '18 at 9:48
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 '18 at 9:45
Yun Chien YenYun Chien Yen
31
edited Nov 29 '18 at 9:48
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 29 '18 at 9:48
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 29 '18 at 9:48
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Nov 29 '18 at 9:45
Yun Chien YenYun Chien Yen
31
asked Nov 29 '18 at 9:45
Yun Chien YenYun Chien Yen
31
asked Nov 29 '18 at 9:45
Yun Chien YenYun Chien Yen
31
31
closed as off-topic by Kavi Rama Murthy, caverac, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Rebellos Nov 29 '18 at 12:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, caverac, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Kavi Rama Murthy, caverac, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Rebellos Nov 29 '18 at 12:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, caverac, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
Markov's inequality applies to all r.v., so it utilizes little characteristics of a specific probability distribution. As a result, the upper bound given by Markov/Chebylshev inequality is often too weak.
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:50
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 29 '18 at 9:50
add a comment |
Markov's inequality applies to all r.v., so it utilizes little characteristics of a specific probability distribution. As a result, the upper bound given by Markov/Chebylshev inequality is often too weak.
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:50
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 29 '18 at 9:50
Markov's inequality applies to all r.v., so it utilizes little characteristics of a specific probability distribution. As a result, the upper bound given by Markov/Chebylshev inequality is often too weak.
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:50
Markov's inequality applies to all r.v., so it utilizes little characteristics of a specific probability distribution. As a result, the upper bound given by Markov/Chebylshev inequality is often too weak.
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:50
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 29 '18 at 9:50
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 29 '18 at 9:50
add a comment |
1 Answer
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Markov's Inequality states that if $X$ is a nonnegative random variable and
$c > 0$, then $$P(X geq c) leq frac{mathbb{E}[X]}{c}.$$
We have $X sim text{Poisson}(1)$, meaning that $mathbb{E}[X] = lambda = 1$. We can set $c = 2$. Then, we obtain the bound
$$P(X geq 2) leq frac{mathbb{E}[X]}{2} = frac{1}{2}.$$
Therefore, we conclude $P(X geq 2)$ will be at most $0.5$.
answered Nov 29 '18 at 11:52
EkeshEkesh
5326
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Markov's Inequality states that if $X$ is a nonnegative random variable and
$c > 0$, then $$P(X geq c) leq frac{mathbb{E}[X]}{c}.$$
We have $X sim text{Poisson}(1)$, meaning that $mathbb{E}[X] = lambda = 1$. We can set $c = 2$. Then, we obtain the bound
$$P(X geq 2) leq frac{mathbb{E}[X]}{2} = frac{1}{2}.$$
Therefore, we conclude $P(X geq 2)$ will be at most $0.5$.
answered Nov 29 '18 at 11:52
EkeshEkesh
5326
add a comment |
Markov's Inequality states that if $X$ is a nonnegative random variable and
$c > 0$, then $$P(X geq c) leq frac{mathbb{E}[X]}{c}.$$
We have $X sim text{Poisson}(1)$, meaning that $mathbb{E}[X] = lambda = 1$. We can set $c = 2$. Then, we obtain the bound
$$P(X geq 2) leq frac{mathbb{E}[X]}{2} = frac{1}{2}.$$
Therefore, we conclude $P(X geq 2)$ will be at most $0.5$.
answered Nov 29 '18 at 11:52
EkeshEkesh
5326
add a comment |
Markov's Inequality states that if $X$ is a nonnegative random variable and
$c > 0$, then $$P(X geq c) leq frac{mathbb{E}[X]}{c}.$$
We have $X sim text{Poisson}(1)$, meaning that $mathbb{E}[X] = lambda = 1$. We can set $c = 2$. Then, we obtain the bound
$$P(X geq 2) leq frac{mathbb{E}[X]}{2} = frac{1}{2}.$$
Therefore, we conclude $P(X geq 2)$ will be at most $0.5$.
answered Nov 29 '18 at 11:52
EkeshEkesh
5326
Markov's Inequality states that if $X$ is a nonnegative random variable and
$c > 0$, then $$P(X geq c) leq frac{mathbb{E}[X]}{c}.$$
We have $X sim text{Poisson}(1)$, meaning that $mathbb{E}[X] = lambda = 1$. We can set $c = 2$. Then, we obtain the bound
$$P(X geq 2) leq frac{mathbb{E}[X]}{2} = frac{1}{2}.$$
Therefore, we conclude $P(X geq 2)$ will be at most $0.5$.
answered Nov 29 '18 at 11:52
EkeshEkesh
5326
answered Nov 29 '18 at 11:52
EkeshEkesh
5326
answered Nov 29 '18 at 11:52
EkeshEkesh
5326
answered Nov 29 '18 at 11:52
EkeshEkesh
5326
5326
add a comment |
add a comment |
Markov's inequality applies to all r.v., so it utilizes little characteristics of a specific probability distribution. As a result, the upper bound given by Markov/Chebylshev inequality is often too weak.
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:50
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 29 '18 at 9:50