Find CDF when knowing PDF, also find E[X]Finding CDF for PDFFind constant $c$ for piecewise continuous random variable pdfGiven a CDF find the PDFHow to find the CDF and PDFFind the CDF from the PDF.Finding probability density function for $Z$ which is a function of $X$ and $Y$Set $Z=X+Y$ and calculate the density function $f_Z$ for Z. Given solution doesn't match mine. (Continuous R.V)Find cdf given pdf.Find the cumulative probability function given a probability density functionProbability of one random variable less than two otherSpiral path quasi-CDF sampling approach question
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Find CDF when knowing PDF, also find E[X]
Finding CDF for PDFFind constant $c$ for piecewise continuous random variable pdfGiven a CDF find the PDFHow to find the CDF and PDFFind the CDF from the PDF.Finding probability density function for $Z$ which is a function of $X$ and $Y$Set $Z=X+Y$ and calculate the density function $f_Z$ for Z. Given solution doesn't match mine. (Continuous R.V)Find cdf given pdf.Find the cumulative probability function given a probability density functionProbability of one random variable less than two otherSpiral path quasi-CDF sampling approach question
$begingroup$
i lose my login information and have to make new account
i apologies for my poor english
i am given a problem as such:
$$ f(x) = begincases
c(1-x^2), & -1<x<0, \ c/x^2, & 1<x<2, \ 0, & textotherwise.
endcases$$
first step is to find c
I calculate c = 6 by following:
$$ int^0_-1 c(1-x^2) , dx + int^2_1 fraccx^2 , dx = 1 $$
is this correct?
2 more part to the problem which trouble me
find CDF $$ F(X) = ? $$
find E(X)
$$ E(x) = int_-infty^inftyxf(x)dx $$
i am trouble by the f(x) not being continue from 0, 1
this is a new problem sort to me
to the one who helps me i am very thankful
probability probability-distributions
edited Dec 7 '14 at 18:13
mathisfun
18112
asked Dec 7 '14 at 17:26
chinpichinpi
1
$endgroup$
add a comment |
$begingroup$
i lose my login information and have to make new account
i apologies for my poor english
i am given a problem as such:
$$ f(x) = begincases
c(1-x^2), & -1<x<0, \ c/x^2, & 1<x<2, \ 0, & textotherwise.
endcases$$
first step is to find c
I calculate c = 6 by following:
$$ int^0_-1 c(1-x^2) , dx + int^2_1 fraccx^2 , dx = 1 $$
is this correct?
2 more part to the problem which trouble me
find CDF $$ F(X) = ? $$
find E(X)
$$ E(x) = int_-infty^inftyxf(x)dx $$
i am trouble by the f(x) not being continue from 0, 1
this is a new problem sort to me
to the one who helps me i am very thankful
probability probability-distributions
edited Dec 7 '14 at 18:13
mathisfun
18112
asked Dec 7 '14 at 17:26
chinpichinpi
1
$endgroup$
1
$begingroup$
No, $displaystyle c=frac67$
$endgroup$
– Extremal
Dec 7 '14 at 17:32
add a comment |
$begingroup$
i lose my login information and have to make new account
i apologies for my poor english
i am given a problem as such:
$$ f(x) = begincases
c(1-x^2), & -1<x<0, \ c/x^2, & 1<x<2, \ 0, & textotherwise.
endcases$$
first step is to find c
I calculate c = 6 by following:
$$ int^0_-1 c(1-x^2) , dx + int^2_1 fraccx^2 , dx = 1 $$
is this correct?
2 more part to the problem which trouble me
find CDF $$ F(X) = ? $$
find E(X)
$$ E(x) = int_-infty^inftyxf(x)dx $$
i am trouble by the f(x) not being continue from 0, 1
this is a new problem sort to me
to the one who helps me i am very thankful
probability probability-distributions
edited Dec 7 '14 at 18:13
mathisfun
18112
asked Dec 7 '14 at 17:26
chinpichinpi
1
$endgroup$
i lose my login information and have to make new account
i apologies for my poor english
i am given a problem as such:
$$ f(x) = begincases
c(1-x^2), & -1<x<0, \ c/x^2, & 1<x<2, \ 0, & textotherwise.
endcases$$
first step is to find c
I calculate c = 6 by following:
$$ int^0_-1 c(1-x^2) , dx + int^2_1 fraccx^2 , dx = 1 $$
is this correct?
2 more part to the problem which trouble me
find CDF $$ F(X) = ? $$
find E(X)
$$ E(x) = int_-infty^inftyxf(x)dx $$
i am trouble by the f(x) not being continue from 0, 1
this is a new problem sort to me
to the one who helps me i am very thankful
probability probability-distributions
probability probability-distributions
edited Dec 7 '14 at 18:13
mathisfun
18112
asked Dec 7 '14 at 17:26
chinpichinpi
1
edited Dec 7 '14 at 18:13
mathisfun
18112
asked Dec 7 '14 at 17:26
chinpichinpi
1
edited Dec 7 '14 at 18:13
mathisfun
18112
edited Dec 7 '14 at 18:13
mathisfun
18112
edited Dec 7 '14 at 18:13
mathisfun
18112
18112
asked Dec 7 '14 at 17:26
chinpichinpi
1
asked Dec 7 '14 at 17:26
chinpichinpi
1
asked Dec 7 '14 at 17:26
chinpichinpi
1
1
1
$begingroup$
No, $displaystyle c=frac67$
$endgroup$
– Extremal
Dec 7 '14 at 17:32
add a comment |
1
$begingroup$
No, $displaystyle c=frac67$
$endgroup$
– Extremal
Dec 7 '14 at 17:32
1
1
$begingroup$
No, $displaystyle c=frac67$
$endgroup$
– Extremal
Dec 7 '14 at 17:32
$begingroup$
No, $displaystyle c=frac67$
$endgroup$
– Extremal
Dec 7 '14 at 17:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You calculated $c$ wrong. It should be $c = frac67$ and then according to:
Finding CDF for PDF
You will get the following CDF:
beginequation
F(x) =
begincases
0, text if x<-1 \
frac17 left(4+6t-2t^3 right), text if x in [-1,0]\
frac47, text if x in [0,1] \
frac47+frac67 left(1-frac1t right) text if x in [1,2] \
1, text if x>2
endcases
endequation
For expectation, you should split your integral in the same way.
$$mathrmE, [X] = int_-1^0 frac67 x (1-x^2) , mathrmd , x + int_1^2 frac67x , mathrmd x = frac67 ln2 - frac314$$
I hope I didn't do a mistake in expectation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 7 '14 at 17:56
mathisfunmathisfun
18112
$endgroup$
$begingroup$
how can this be? this gives F(0.999) = 1.14 which > 1?
$endgroup$
– chinpi
Dec 7 '14 at 18:35
$begingroup$
@chinpi sure, made some correction
$endgroup$
– mathisfun
Dec 7 '14 at 19:00
1
$begingroup$
$E[x]$ is wrong. Please note that the upper bound on the first integral to calculate $E[x]$ should be $0$ instead of $1$.
$endgroup$
– Math-fun
Dec 7 '14 at 19:05
$begingroup$
@Mehdi sure, let me correct this
$endgroup$
– mathisfun
Dec 7 '14 at 19:06
$begingroup$
this is now correct, and i am happy i thank you kindly
$endgroup$
– chinpi
Dec 7 '14 at 19:11
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
You calculated $c$ wrong. It should be $c = frac67$ and then according to:
Finding CDF for PDF
You will get the following CDF:
beginequation
F(x) =
begincases
0, text if x<-1 \
frac17 left(4+6t-2t^3 right), text if x in [-1,0]\
frac47, text if x in [0,1] \
frac47+frac67 left(1-frac1t right) text if x in [1,2] \
1, text if x>2
endcases
endequation
For expectation, you should split your integral in the same way.
$$mathrmE, [X] = int_-1^0 frac67 x (1-x^2) , mathrmd , x + int_1^2 frac67x , mathrmd x = frac67 ln2 - frac314$$
I hope I didn't do a mistake in expectation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 7 '14 at 17:56
mathisfunmathisfun
18112
$endgroup$
$begingroup$
how can this be? this gives F(0.999) = 1.14 which > 1?
$endgroup$
– chinpi
Dec 7 '14 at 18:35
$begingroup$
@chinpi sure, made some correction
$endgroup$
– mathisfun
Dec 7 '14 at 19:00
1
$begingroup$
$E[x]$ is wrong. Please note that the upper bound on the first integral to calculate $E[x]$ should be $0$ instead of $1$.
$endgroup$
– Math-fun
Dec 7 '14 at 19:05
$begingroup$
@Mehdi sure, let me correct this
$endgroup$
– mathisfun
Dec 7 '14 at 19:06
$begingroup$
this is now correct, and i am happy i thank you kindly
$endgroup$
– chinpi
Dec 7 '14 at 19:11
add a comment |
$begingroup$
You calculated $c$ wrong. It should be $c = frac67$ and then according to:
Finding CDF for PDF
You will get the following CDF:
beginequation
F(x) =
begincases
0, text if x<-1 \
frac17 left(4+6t-2t^3 right), text if x in [-1,0]\
frac47, text if x in [0,1] \
frac47+frac67 left(1-frac1t right) text if x in [1,2] \
1, text if x>2
endcases
endequation
For expectation, you should split your integral in the same way.
$$mathrmE, [X] = int_-1^0 frac67 x (1-x^2) , mathrmd , x + int_1^2 frac67x , mathrmd x = frac67 ln2 - frac314$$
I hope I didn't do a mistake in expectation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 7 '14 at 17:56
mathisfunmathisfun
18112
$endgroup$
$begingroup$
how can this be? this gives F(0.999) = 1.14 which > 1?
$endgroup$
– chinpi
Dec 7 '14 at 18:35
$begingroup$
@chinpi sure, made some correction
$endgroup$
– mathisfun
Dec 7 '14 at 19:00
1
$begingroup$
$E[x]$ is wrong. Please note that the upper bound on the first integral to calculate $E[x]$ should be $0$ instead of $1$.
$endgroup$
– Math-fun
Dec 7 '14 at 19:05
$begingroup$
@Mehdi sure, let me correct this
$endgroup$
– mathisfun
Dec 7 '14 at 19:06
$begingroup$
this is now correct, and i am happy i thank you kindly
$endgroup$
– chinpi
Dec 7 '14 at 19:11
add a comment |
$begingroup$
You calculated $c$ wrong. It should be $c = frac67$ and then according to:
Finding CDF for PDF
You will get the following CDF:
beginequation
F(x) =
begincases
0, text if x<-1 \
frac17 left(4+6t-2t^3 right), text if x in [-1,0]\
frac47, text if x in [0,1] \
frac47+frac67 left(1-frac1t right) text if x in [1,2] \
1, text if x>2
endcases
endequation
For expectation, you should split your integral in the same way.
$$mathrmE, [X] = int_-1^0 frac67 x (1-x^2) , mathrmd , x + int_1^2 frac67x , mathrmd x = frac67 ln2 - frac314$$
I hope I didn't do a mistake in expectation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 7 '14 at 17:56
mathisfunmathisfun
18112
$endgroup$
You calculated $c$ wrong. It should be $c = frac67$ and then according to:
Finding CDF for PDF
You will get the following CDF:
beginequation
F(x) =
begincases
0, text if x<-1 \
frac17 left(4+6t-2t^3 right), text if x in [-1,0]\
frac47, text if x in [0,1] \
frac47+frac67 left(1-frac1t right) text if x in [1,2] \
1, text if x>2
endcases
endequation
For expectation, you should split your integral in the same way.
$$mathrmE, [X] = int_-1^0 frac67 x (1-x^2) , mathrmd , x + int_1^2 frac67x , mathrmd x = frac67 ln2 - frac314$$
I hope I didn't do a mistake in expectation.
edited Apr 13 '17 at 12:20
Community♦
1
answered Dec 7 '14 at 17:56
mathisfunmathisfun
18112
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Dec 7 '14 at 17:56
mathisfunmathisfun
18112
answered Dec 7 '14 at 17:56
mathisfunmathisfun
18112
answered Dec 7 '14 at 17:56
mathisfunmathisfun
18112
18112
$begingroup$
how can this be? this gives F(0.999) = 1.14 which > 1?
$endgroup$
– chinpi
Dec 7 '14 at 18:35
$begingroup$
@chinpi sure, made some correction
$endgroup$
– mathisfun
Dec 7 '14 at 19:00
1
$begingroup$
$E[x]$ is wrong. Please note that the upper bound on the first integral to calculate $E[x]$ should be $0$ instead of $1$.
$endgroup$
– Math-fun
Dec 7 '14 at 19:05
$begingroup$
@Mehdi sure, let me correct this
$endgroup$
– mathisfun
Dec 7 '14 at 19:06
$begingroup$
this is now correct, and i am happy i thank you kindly
$endgroup$
– chinpi
Dec 7 '14 at 19:11
add a comment |
$begingroup$
how can this be? this gives F(0.999) = 1.14 which > 1?
$endgroup$
– chinpi
Dec 7 '14 at 18:35
$begingroup$
@chinpi sure, made some correction
$endgroup$
– mathisfun
Dec 7 '14 at 19:00
1
$begingroup$
$E[x]$ is wrong. Please note that the upper bound on the first integral to calculate $E[x]$ should be $0$ instead of $1$.
$endgroup$
– Math-fun
Dec 7 '14 at 19:05
$begingroup$
@Mehdi sure, let me correct this
$endgroup$
– mathisfun
Dec 7 '14 at 19:06
$begingroup$
this is now correct, and i am happy i thank you kindly
$endgroup$
– chinpi
Dec 7 '14 at 19:11
$begingroup$
how can this be? this gives F(0.999) = 1.14 which > 1?
$endgroup$
– chinpi
Dec 7 '14 at 18:35
$begingroup$
how can this be? this gives F(0.999) = 1.14 which > 1?
$endgroup$
– chinpi
Dec 7 '14 at 18:35
$begingroup$
@chinpi sure, made some correction
$endgroup$
– mathisfun
Dec 7 '14 at 19:00
$begingroup$
@chinpi sure, made some correction
$endgroup$
– mathisfun
Dec 7 '14 at 19:00
1
1
$begingroup$
$E[x]$ is wrong. Please note that the upper bound on the first integral to calculate $E[x]$ should be $0$ instead of $1$.
$endgroup$
– Math-fun
Dec 7 '14 at 19:05
$begingroup$
$E[x]$ is wrong. Please note that the upper bound on the first integral to calculate $E[x]$ should be $0$ instead of $1$.
$endgroup$
– Math-fun
Dec 7 '14 at 19:05
$begingroup$
@Mehdi sure, let me correct this
$endgroup$
– mathisfun
Dec 7 '14 at 19:06
$begingroup$
@Mehdi sure, let me correct this
$endgroup$
– mathisfun
Dec 7 '14 at 19:06
$begingroup$
this is now correct, and i am happy i thank you kindly
$endgroup$
– chinpi
Dec 7 '14 at 19:11
$begingroup$
this is now correct, and i am happy i thank you kindly
$endgroup$
– chinpi
Dec 7 '14 at 19:11
add a comment |
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$begingroup$
No, $displaystyle c=frac67$
$endgroup$
– Extremal
Dec 7 '14 at 17:32