Yhyjyk

Multi tool use

What is the command to reset a PC without deleting any files

declaring a variable twice in IIFE

If Manufacturer spice model and Datasheet give different values which should I use?

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

Is it possible to do 50 km distance without any previous training?

What does "enim et" mean?

How can the DM most effectively choose 1 out of an odd number of players to be targeted by an attack or effect?

Why is an old chain unsafe?

How is the relation "the smallest element is the same" reflexive?

What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?

Is it possible to make sharp wind that can cut stuff from afar?

Infinite past with a beginning?

Can I make popcorn with any corn?

The magic money tree problem

Is it legal to have the "// (c) 2019 John Smith" header in all files when there are hundreds of contributors?

Motorized valve interfering with button?

Non-Jewish family in an Orthodox Jewish Wedding

Should I join an office cleaning event for free?

Is Social Media Science Fiction?

Why is this code 6.5x slower with optimizations enabled?

Prevent a directory in /tmp from being deleted

Example of a relative pronoun

Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?

A Journey Through Space and Time

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)

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

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 | 

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

votes

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 | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "65"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

EBusch is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f401726%2fmean-and-variance-of-continuous-random-variable%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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