How to find probability
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Consider this question:
Mean of $X$ is 20 and upper bound is 50. If $X$ changes daily with mean 0 and variance 1, independently of the other days. If today $X=40$ what can you say about the probability that $X$ will be between 35 and 45 after 5 days.
ChebyChev is supposed to be used for this. I am unsure of how to use it. If the mean is 20 and the value is 40 then tomorrow mean will be 20 and variance would be increased by 1. So, whatever variance is today it will be 5 times, 5 days from now. So, P(35 <= X <= 45) = P(X - 20 >= 5*var) = 1/25.
Is this correct?
probability probability-theory
asked Nov 15 at 16:27
puffles
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Looking for an answer drawing from credible and/or official sources.
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Consider this question:
Mean of $X$ is 20 and upper bound is 50. If $X$ changes daily with mean 0 and variance 1, independently of the other days. If today $X=40$ what can you say about the probability that $X$ will be between 35 and 45 after 5 days.
ChebyChev is supposed to be used for this. I am unsure of how to use it. If the mean is 20 and the value is 40 then tomorrow mean will be 20 and variance would be increased by 1. So, whatever variance is today it will be 5 times, 5 days from now. So, P(35 <= X <= 45) = P(X - 20 >= 5*var) = 1/25.
Is this correct?
probability probability-theory
asked Nov 15 at 16:27
puffles
669
This question has an open bounty worth +50
reputation from puffles ending in 3 days.
Looking for an answer drawing from credible and/or official sources.
"Looks like standard normal RV" Yes, somehow it does. But it is not mentioned in the text.
– callculus
Nov 15 at 17:18
Can I solve it using ChebyChev or Chernoff? What would be the steps taken further?
– puffles
Nov 16 at 6:20
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider this question:
Mean of $X$ is 20 and upper bound is 50. If $X$ changes daily with mean 0 and variance 1, independently of the other days. If today $X=40$ what can you say about the probability that $X$ will be between 35 and 45 after 5 days.
ChebyChev is supposed to be used for this. I am unsure of how to use it. If the mean is 20 and the value is 40 then tomorrow mean will be 20 and variance would be increased by 1. So, whatever variance is today it will be 5 times, 5 days from now. So, P(35 <= X <= 45) = P(X - 20 >= 5*var) = 1/25.
Is this correct?
probability probability-theory
asked Nov 15 at 16:27
puffles
669
Consider this question:
Mean of $X$ is 20 and upper bound is 50. If $X$ changes daily with mean 0 and variance 1, independently of the other days. If today $X=40$ what can you say about the probability that $X$ will be between 35 and 45 after 5 days.
ChebyChev is supposed to be used for this. I am unsure of how to use it. If the mean is 20 and the value is 40 then tomorrow mean will be 20 and variance would be increased by 1. So, whatever variance is today it will be 5 times, 5 days from now. So, P(35 <= X <= 45) = P(X - 20 >= 5*var) = 1/25.
Is this correct?
probability probability-theory
probability probability-theory
asked Nov 15 at 16:27
puffles
669
asked Nov 15 at 16:27
puffles
669
edited Nov 18 at 10:24
asked Nov 15 at 16:27
puffles
669
asked Nov 15 at 16:27
puffles
669
asked Nov 15 at 16:27
puffles
669
669
This question has an open bounty worth +50
reputation from puffles ending in 3 days.
Looking for an answer drawing from credible and/or official sources.
This question has an open bounty worth +50
reputation from puffles ending in 3 days.
Looking for an answer drawing from credible and/or official sources.
"Looks like standard normal RV" Yes, somehow it does. But it is not mentioned in the text.
– callculus
Nov 15 at 17:18
Can I solve it using ChebyChev or Chernoff? What would be the steps taken further?
– puffles
Nov 16 at 6:20
add a comment |
"Looks like standard normal RV" Yes, somehow it does. But it is not mentioned in the text.
– callculus
Nov 15 at 17:18
Can I solve it using ChebyChev or Chernoff? What would be the steps taken further?
– puffles
Nov 16 at 6:20
"Looks like standard normal RV" Yes, somehow it does. But it is not mentioned in the text.
– callculus
Nov 15 at 17:18
"Looks like standard normal RV" Yes, somehow it does. But it is not mentioned in the text.
– callculus
Nov 15 at 17:18
Can I solve it using ChebyChev or Chernoff? What would be the steps taken further?
– puffles
Nov 16 at 6:20
Can I solve it using ChebyChev or Chernoff? What would be the steps taken further?
– puffles
Nov 16 at 6:20
add a comment |
2 Answers
2
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oldest
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up vote
0
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Well, I think this question want you to point out that the probability of X = 40 in today does not affect the probability of X in the next 5 days. Because it is independently of the other days.
So, if the distribution of X is standard normal distribution, then you can compute the $P(35<=X<=45)$ directly.
answered Nov 15 at 18:16
AnNg
375
New contributor
AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The changes from day to day are independent, which does not make the values on each day independent.
– wnoise
Nov 15 at 18:41
Do you think it canbe solved using any inequality (Markov, chebychev, chernoff)
– puffles
Nov 16 at 6:20
add a comment |
up vote
0
down vote
The sum of independent Gaussian RVs is a Gaussian RV with a variance that is the sum of the individual variances, and a mean that is the sum of the means. From this, the PDF of values on the last day is immediately available.
answered Nov 15 at 18:43
wnoise
1,4331017
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Well, I think this question want you to point out that the probability of X = 40 in today does not affect the probability of X in the next 5 days. Because it is independently of the other days.
So, if the distribution of X is standard normal distribution, then you can compute the $P(35<=X<=45)$ directly.
answered Nov 15 at 18:16
AnNg
375
New contributor
AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The changes from day to day are independent, which does not make the values on each day independent.
– wnoise
Nov 15 at 18:41
Do you think it canbe solved using any inequality (Markov, chebychev, chernoff)
– puffles
Nov 16 at 6:20
add a comment |
up vote
0
down vote
Well, I think this question want you to point out that the probability of X = 40 in today does not affect the probability of X in the next 5 days. Because it is independently of the other days.
So, if the distribution of X is standard normal distribution, then you can compute the $P(35<=X<=45)$ directly.
answered Nov 15 at 18:16
AnNg
375
New contributor
AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The changes from day to day are independent, which does not make the values on each day independent.
– wnoise
Nov 15 at 18:41
Do you think it canbe solved using any inequality (Markov, chebychev, chernoff)
– puffles
Nov 16 at 6:20
add a comment |
up vote
0
down vote
up vote
0
down vote
Well, I think this question want you to point out that the probability of X = 40 in today does not affect the probability of X in the next 5 days. Because it is independently of the other days.
So, if the distribution of X is standard normal distribution, then you can compute the $P(35<=X<=45)$ directly.
answered Nov 15 at 18:16
AnNg
375
New contributor
AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Well, I think this question want you to point out that the probability of X = 40 in today does not affect the probability of X in the next 5 days. Because it is independently of the other days.
So, if the distribution of X is standard normal distribution, then you can compute the $P(35<=X<=45)$ directly.
answered Nov 15 at 18:16
AnNg
375
New contributor
AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 15 at 18:16
AnNg
375
New contributor
AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 15 at 18:16
AnNg
375
answered Nov 15 at 18:16
AnNg
375
375
New contributor
AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The changes from day to day are independent, which does not make the values on each day independent.
– wnoise
Nov 15 at 18:41
Do you think it canbe solved using any inequality (Markov, chebychev, chernoff)
– puffles
Nov 16 at 6:20
add a comment |
The changes from day to day are independent, which does not make the values on each day independent.
– wnoise
Nov 15 at 18:41
Do you think it canbe solved using any inequality (Markov, chebychev, chernoff)
– puffles
Nov 16 at 6:20
The changes from day to day are independent, which does not make the values on each day independent.
– wnoise
Nov 15 at 18:41
The changes from day to day are independent, which does not make the values on each day independent.
– wnoise
Nov 15 at 18:41
Do you think it canbe solved using any inequality (Markov, chebychev, chernoff)
– puffles
Nov 16 at 6:20
Do you think it canbe solved using any inequality (Markov, chebychev, chernoff)
– puffles
Nov 16 at 6:20
add a comment |
up vote
0
down vote
The sum of independent Gaussian RVs is a Gaussian RV with a variance that is the sum of the individual variances, and a mean that is the sum of the means. From this, the PDF of values on the last day is immediately available.
answered Nov 15 at 18:43
wnoise
1,4331017
add a comment |
up vote
0
down vote
The sum of independent Gaussian RVs is a Gaussian RV with a variance that is the sum of the individual variances, and a mean that is the sum of the means. From this, the PDF of values on the last day is immediately available.
answered Nov 15 at 18:43
wnoise
1,4331017
add a comment |
up vote
0
down vote
up vote
0
down vote
The sum of independent Gaussian RVs is a Gaussian RV with a variance that is the sum of the individual variances, and a mean that is the sum of the means. From this, the PDF of values on the last day is immediately available.
answered Nov 15 at 18:43
wnoise
1,4331017
The sum of independent Gaussian RVs is a Gaussian RV with a variance that is the sum of the individual variances, and a mean that is the sum of the means. From this, the PDF of values on the last day is immediately available.
answered Nov 15 at 18:43
wnoise
1,4331017
answered Nov 15 at 18:43
wnoise
1,4331017
answered Nov 15 at 18:43
wnoise
1,4331017
answered Nov 15 at 18:43
wnoise
1,4331017
1,4331017
add a comment |
add a comment |
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"Looks like standard normal RV" Yes, somehow it does. But it is not mentioned in the text.
– callculus
Nov 15 at 17:18
Can I solve it using ChebyChev or Chernoff? What would be the steps taken further?
– puffles
Nov 16 at 6:20