Mean and Variance of Continuous Random Variable
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I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac{1}{(1+e^{-y})}$.
Part a is asking for the pdf which I found to be $frac{e^y}{(e^y+1)^2}$.
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 $frac{ye^y}{(e^y+1)^2}$. I'm not sure where I'm going wrong with this problem?
variance mean
edited 3 hours ago
Noah
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asked 5 hours ago
EBuschEBusch
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add a comment |
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I have a problem on my homework about the continuous random variable $y$ where the cdf is $F(y)=frac{1}{(1+e^{-y})}$.
Part a is asking for the pdf which I found to be $frac{e^y}{(e^y+1)^2}$.
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 $frac{ye^y}{(e^y+1)^2}$. I'm not sure where I'm going wrong with this problem?
variance mean
edited 3 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 $F(y)=frac{1}{(1+e^{-y})}$.
Part a is asking for the pdf which I found to be $frac{e^y}{(e^y+1)^2}$.
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 $frac{ye^y}{(e^y+1)^2}$. I'm not sure where I'm going wrong with this problem?
variance mean
edited 3 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 $F(y)=frac{1}{(1+e^{-y})}$.
Part a is asking for the pdf which I found to be $frac{e^y}{(e^y+1)^2}$.
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 $frac{ye^y}{(e^y+1)^2}$. I'm not sure where I'm going wrong with this problem?
variance mean
variance mean
edited 3 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 3 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 3 hours ago
Noah
3,6011417
edited 3 hours ago
Noah
3,6011417
edited 3 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.
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2 Answers
2
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$begingroup$
What makes you think you did something wrong?
begin{align}
& Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
= {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
end{align}
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:
begin{align}
f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
end{align}
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
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add a comment |
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Comment:
Setting what I take to be your CDF equal to $U sim mathsf{Unif}(0,1),$ 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
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add a comment |
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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?
begin{align}
& Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
= {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
end{align}
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:
begin{align}
f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
end{align}
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?
begin{align}
& Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
= {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
end{align}
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:
begin{align}
f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
end{align}
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?
begin{align}
& Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
= {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
end{align}
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:
begin{align}
f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
end{align}
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?
begin{align}
& Pr(Yle y) = F(y) = frac 1 {1+e^{-y}} \[10pt]
text{and } & Pr(Yge -y) = 1-F(-y) = 1- frac 1 {1+e^y} \[8pt]
= {} & frac{e^y}{1+e^y} = frac{e^ycdot e^{-y}}{(1+e^y)cdot e^{-y}} = frac 1 {e^{-y}+1},
end{align}
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:
begin{align}
f(y) & = frac{e^y}{(1+e^y)^2}. \[12pt]
f(-y) & = frac{e^{-y}}{(1+e^{-y})^2} = frac{e^{-y}cdotleft( e^y right)^2}{Big((1+e^{-y}) cdot e^y Big)^2} = frac{e^y}{(e^y+1)^2} = f(y).
end{align}
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 $U sim mathsf{Unif}(0,1),$ 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 $U sim mathsf{Unif}(0,1),$ 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 $U sim mathsf{Unif}(0,1),$ 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 $U sim mathsf{Unif}(0,1),$ 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 28 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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