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2

$begingroup$

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

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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.
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add a comment | 

2

$begingroup$

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

share|cite|improve this question

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 $F(y)=frac1(1+e^-y)$.

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

share|cite|improve this question

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.

$endgroup$

share|cite|improve this question

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.

share|cite|improve this question

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

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

2 Answers
2

active

oldest

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1

$begingroup$

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 $operatorname E(|Y|) < +infty.$

share|cite|improve this answer

answered 2 hours ago

Michael HardyMichael Hardy

3,9951430

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

edited 31 mins ago

answered 4 hours ago

BruceETBruceET

6,3531721

$endgroup$

add a comment | 

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1

$begingroup$

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 $operatorname E(|Y|) < +infty.$

share|cite|improve this answer

answered 2 hours ago

Michael HardyMichael Hardy

3,9951430

$endgroup$

add a comment | 

1

$begingroup$

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 $operatorname E(|Y|) < +infty.$

share|cite|improve this answer

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 $operatorname E(|Y|) < +infty.$

share|cite|improve this answer

answered 2 hours ago

Michael HardyMichael Hardy

3,9951430

$endgroup$

share|cite|improve this answer

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

0

$begingroup$

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.

share|cite|improve this answer

edited 31 mins ago

answered 4 hours ago

BruceETBruceET

6,3531721

$endgroup$

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

edited 31 mins ago

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.

share|cite|improve this answer

edited 31 mins ago

answered 4 hours ago

BruceETBruceET

6,3531721

$endgroup$

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")

share|cite|improve this answer

edited 31 mins ago

answered 4 hours ago

BruceETBruceET

6,3531721

answered 4 hours ago

BruceETBruceET

6,3531721

answered 4 hours ago

BruceETBruceET

6,3531721