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Mean and Variance of Continuous Random Variable
Mean and variance of call center dataPoisson random variable- varianceHow to obtain variance of a random variable that depends on a hypergeometric variable?Variance reduction technique in Monte Carlo integrationVariance of a continuous uniformly distributed random variableWeighted sample mean and variance - asking for references and detailsCan the variance of a continuous random variable with known distribution be impossible to find?Mean and variance of the maximum of a random number of Uniform variablesApproximating the expected value and variance of the function of a (continuous univariate) random variableMean of maximum of exponential random variables (independent but not identical)
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$begingroup$
I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac1(1+e^-y)$.
Part a is asking for the pdf which I found to be $frace^y(e^y+1)^2$.
Part b asks for the mean and variance of $y$ but when I tried to find the $E(y)$, I got zero with the integral from $-infty$ to $infty$ of $fracye^y(e^y+1)^2$. I'm not sure where I'm going wrong with this problem?
variance mean
edited 2 hours ago
Noah
3,6011417
asked 5 hours ago
EBuschEBusch
112
New contributor
EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac1(1+e^-y)$.
Part a is asking for the pdf which I found to be $frace^y(e^y+1)^2$.
Part b asks for the mean and variance of $y$ but when I tried to find the $E(y)$, I got zero with the integral from $-infty$ to $infty$ of $fracye^y(e^y+1)^2$. I'm not sure where I'm going wrong with this problem?
variance mean
edited 2 hours ago
Noah
3,6011417
asked 5 hours ago
EBuschEBusch
112
New contributor
EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac1(1+e^-y)$.
Part a is asking for the pdf which I found to be $frace^y(e^y+1)^2$.
Part b asks for the mean and variance of $y$ but when I tried to find the $E(y)$, I got zero with the integral from $-infty$ to $infty$ of $fracye^y(e^y+1)^2$. I'm not sure where I'm going wrong with this problem?
variance mean
edited 2 hours ago
Noah
3,6011417
asked 5 hours ago
EBuschEBusch
112
New contributor
EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac1(1+e^-y)$.
Part a is asking for the pdf which I found to be $frace^y(e^y+1)^2$.
Part b asks for the mean and variance of $y$ but when I tried to find the $E(y)$, I got zero with the integral from $-infty$ to $infty$ of $fracye^y(e^y+1)^2$. I'm not sure where I'm going wrong with this problem?
variance mean
variance mean
edited 2 hours ago
Noah
3,6011417
asked 5 hours ago
EBuschEBusch
112
New contributor
EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Noah
3,6011417
asked 5 hours ago
EBuschEBusch
112
New contributor
EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Noah
3,6011417
edited 2 hours ago
Noah
3,6011417
edited 2 hours ago
Noah
3,6011417
3,6011417
asked 5 hours ago
EBuschEBusch
112
New contributor
EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
EBuschEBusch
112
asked 5 hours ago
EBuschEBusch
112
112
New contributor
EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
EBusch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
What makes you think you did something wrong?
beginalign
& Pr(Yle y) = F(y) = frac 1 1+e^-y \[10pt]
textand & Pr(Yge -y) = 1-F(-y) = 1- frac 1 1+e^y \[8pt]
= & frace^y1+e^y = frace^ycdot e^-y(1+e^y)cdot e^-y = frac 1 e^-y+1,
endalign
and therefore
$$
Pr(Yle y) = Pr(Y ge -y).
$$
So this distribution is symmetric about $0.$
Therefore, if the expected value exists, it is $0.$
You can also show that the density function is an even function:
beginalign
f(y) & = frace^y(1+e^y)^2. \[12pt]
f(-y) & = frace^-y(1+e^-y)^2 = frace^-ycdotleft( e^y right)^2Big((1+e^-y) cdot e^y Big)^2 = frace^y(e^y+1)^2 = f(y).
endalign
Since the density is an even function, the expected value must be $0$ if it exists.
The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$
answered 2 hours ago
Michael HardyMichael Hardy
3,9951430
$endgroup$
add a comment |
$begingroup$
Comment:
Setting what I take to be your CDF equal to $U sim mathsfUnif(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
observations as shown below.
Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
seems to fit pretty well.
set.seed(1019) # for reproducibility
u = runif(10^6); x = -log(1/u - 1)
hist(x, prob=T, br=100, col="skyblue2")
curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")
I don't pretend that this is a 'worked answer' to your problem, but
I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.
answered 4 hours ago
BruceETBruceET
6,3531721
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What makes you think you did something wrong?
beginalign
& Pr(Yle y) = F(y) = frac 1 1+e^-y \[10pt]
textand & Pr(Yge -y) = 1-F(-y) = 1- frac 1 1+e^y \[8pt]
= & frace^y1+e^y = frace^ycdot e^-y(1+e^y)cdot e^-y = frac 1 e^-y+1,
endalign
and therefore
$$
Pr(Yle y) = Pr(Y ge -y).
$$
So this distribution is symmetric about $0.$
Therefore, if the expected value exists, it is $0.$
You can also show that the density function is an even function:
beginalign
f(y) & = frace^y(1+e^y)^2. \[12pt]
f(-y) & = frace^-y(1+e^-y)^2 = frace^-ycdotleft( e^y right)^2Big((1+e^-y) cdot e^y Big)^2 = frace^y(e^y+1)^2 = f(y).
endalign
Since the density is an even function, the expected value must be $0$ if it exists.
The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$
answered 2 hours ago
Michael HardyMichael Hardy
3,9951430
$endgroup$
add a comment |
$begingroup$
What makes you think you did something wrong?
beginalign
& Pr(Yle y) = F(y) = frac 1 1+e^-y \[10pt]
textand & Pr(Yge -y) = 1-F(-y) = 1- frac 1 1+e^y \[8pt]
= & frace^y1+e^y = frace^ycdot e^-y(1+e^y)cdot e^-y = frac 1 e^-y+1,
endalign
and therefore
$$
Pr(Yle y) = Pr(Y ge -y).
$$
So this distribution is symmetric about $0.$
Therefore, if the expected value exists, it is $0.$
You can also show that the density function is an even function:
beginalign
f(y) & = frace^y(1+e^y)^2. \[12pt]
f(-y) & = frace^-y(1+e^-y)^2 = frace^-ycdotleft( e^y right)^2Big((1+e^-y) cdot e^y Big)^2 = frace^y(e^y+1)^2 = f(y).
endalign
Since the density is an even function, the expected value must be $0$ if it exists.
The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$
answered 2 hours ago
Michael HardyMichael Hardy
3,9951430
$endgroup$
add a comment |
$begingroup$
What makes you think you did something wrong?
beginalign
& Pr(Yle y) = F(y) = frac 1 1+e^-y \[10pt]
textand & Pr(Yge -y) = 1-F(-y) = 1- frac 1 1+e^y \[8pt]
= & frace^y1+e^y = frace^ycdot e^-y(1+e^y)cdot e^-y = frac 1 e^-y+1,
endalign
and therefore
$$
Pr(Yle y) = Pr(Y ge -y).
$$
So this distribution is symmetric about $0.$
Therefore, if the expected value exists, it is $0.$
You can also show that the density function is an even function:
beginalign
f(y) & = frace^y(1+e^y)^2. \[12pt]
f(-y) & = frace^-y(1+e^-y)^2 = frace^-ycdotleft( e^y right)^2Big((1+e^-y) cdot e^y Big)^2 = frace^y(e^y+1)^2 = f(y).
endalign
Since the density is an even function, the expected value must be $0$ if it exists.
The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$
answered 2 hours ago
Michael HardyMichael Hardy
3,9951430
$endgroup$
What makes you think you did something wrong?
beginalign
& Pr(Yle y) = F(y) = frac 1 1+e^-y \[10pt]
textand & Pr(Yge -y) = 1-F(-y) = 1- frac 1 1+e^y \[8pt]
= & frace^y1+e^y = frace^ycdot e^-y(1+e^y)cdot e^-y = frac 1 e^-y+1,
endalign
and therefore
$$
Pr(Yle y) = Pr(Y ge -y).
$$
So this distribution is symmetric about $0.$
Therefore, if the expected value exists, it is $0.$
You can also show that the density function is an even function:
beginalign
f(y) & = frace^y(1+e^y)^2. \[12pt]
f(-y) & = frace^-y(1+e^-y)^2 = frace^-ycdotleft( e^y right)^2Big((1+e^-y) cdot e^y Big)^2 = frace^y(e^y+1)^2 = f(y).
endalign
Since the density is an even function, the expected value must be $0$ if it exists.
The expected value $operatorname E(Y)$ exists if $operatorname E(|Y|) < +infty.$
answered 2 hours ago
Michael HardyMichael Hardy
3,9951430
answered 2 hours ago
Michael HardyMichael Hardy
3,9951430
answered 2 hours ago
Michael HardyMichael Hardy
3,9951430
answered 2 hours ago
Michael HardyMichael Hardy
3,9951430
3,9951430
add a comment |
add a comment |
$begingroup$
Comment:
Setting what I take to be your CDF equal to $U sim mathsfUnif(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
observations as shown below.
Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
seems to fit pretty well.
set.seed(1019) # for reproducibility
u = runif(10^6); x = -log(1/u - 1)
hist(x, prob=T, br=100, col="skyblue2")
curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")
I don't pretend that this is a 'worked answer' to your problem, but
I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.
answered 4 hours ago
BruceETBruceET
6,3531721
$endgroup$
add a comment |
$begingroup$
Comment:
Setting what I take to be your CDF equal to $U sim mathsfUnif(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
observations as shown below.
Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
seems to fit pretty well.
set.seed(1019) # for reproducibility
u = runif(10^6); x = -log(1/u - 1)
hist(x, prob=T, br=100, col="skyblue2")
curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")
I don't pretend that this is a 'worked answer' to your problem, but
I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.
answered 4 hours ago
BruceETBruceET
6,3531721
$endgroup$
add a comment |
$begingroup$
Comment:
Setting what I take to be your CDF equal to $U sim mathsfUnif(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
observations as shown below.
Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
seems to fit pretty well.
set.seed(1019) # for reproducibility
u = runif(10^6); x = -log(1/u - 1)
hist(x, prob=T, br=100, col="skyblue2")
curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")
I don't pretend that this is a 'worked answer' to your problem, but
I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.
answered 4 hours ago
BruceETBruceET
6,3531721
$endgroup$
Comment:
Setting what I take to be your CDF equal to $U sim mathsfUnif(0,1),$ and solving for the quantile function (inverse CDF) in terms of $U,$ I simulate a sample of a million
observations as shown below.
Then, when I plot one possible interpretation of your PDF through the histogram of the large sample, that density function
seems to fit pretty well.
set.seed(1019) # for reproducibility
u = runif(10^6); x = -log(1/u - 1)
hist(x, prob=T, br=100, col="skyblue2")
curve(exp(x)/(exp(x)+1)^2, -10, 10, add=T, lwd=2, col="red")
I don't pretend that this is a 'worked answer' to your problem, but
I hope it may give you enough clues to improve the version of the problem you posted and to finish the problem on your own.
answered 4 hours ago
BruceETBruceET
6,3531721
edited 20 mins ago
answered 4 hours ago
BruceETBruceET
6,3531721
answered 4 hours ago
BruceETBruceET
6,3531721
answered 4 hours ago
BruceETBruceET
6,3531721
6,3531721
add a comment |
add a comment |
EBusch is a new contributor. Be nice, and check out our Code of Conduct.
EBusch is a new contributor. Be nice, and check out our Code of Conduct.
EBusch is a new contributor. Be nice, and check out our Code of Conduct.
EBusch is a new contributor. Be nice, and check out our Code of Conduct.
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