Dtjytyk

A workplace installs custom certificates on personal devices, can this be used to decrypt HTTPS traffic?

Have I saved too much for retirement so far?

Hostile work environment after whistle-blowing on coworker and our boss. What do I do?

What does the "3am" section means in manpages?

What was required to accept "troll"?

What do you call the infoboxes with text and sometimes images on the side of a page we find in textbooks?

Reply ‘no position’ while the job posting is still there (‘HiWi’ position in Germany)

Can somebody explain Brexit in a few child-proof sentences?

Giant Toughroad SLR 2 for 200 miles in two days, will it make it?

What (else) happened July 1st 1858 in London?

Should my PhD thesis be submitted under my legal name?

Visiting the UK as unmarried couple

Is there an Impartial Brexit Deal comparison site?

Who must act to prevent Brexit on March 29th?

Latex for-and in equation

I'm in charge of equipment buying but no one's ever happy with what I choose. How to fix this?

Can I create an upright 7-foot × 5-foot wall with the Minor Illusion spell?

Partial sums of primes

Lightning Web Component - do I need to track changes for every single input field in a form

Is there an wasy way to program in Tikz something like the one in the image?

"lassen" in meaning "sich fassen"

Would it be legal for a US State to ban exports of a natural resource?

Why is delta-v is the most useful quantity for planning space travel?

What is the term when two people sing in harmony, but they aren't singing the same notes?

Mixed random variable (Exponential distribution mixed with binomial distribution)

Sum of exponential random variable with different meansExponential Distribution, Statistics.Exponential distribution of random variableFailure time and exponential distributionDensity of the Sum of Two Exponential Random VariableRandom rate of exponential distributionTransformation of Random Variable $Y = -2ln X$exponential distribution of an exponential variableRandom variable with exponential distribution.Binomial distribution with random variable parameter

1

$begingroup$

Let $X$ be an exponential distribution of parameter $1$.

And let $Y$ be the binomial distribution of parameter $n$ and $p=1/2$.

We define $Z=frac{X}{Y+1}$. I'm trying to find the distribution of $Z$.

This is what I did :

Let $t>0$ we have $F(Z<t)=F(X<t(Y+1))=int_{0}^{t(Y+1)}exp(-x)dx=1-exp(-t(Y+1))$

and for $t<0$ we have $F(Z<t)=0.$

probability probability-theory random-variables

share|cite|improve this question

asked Mar 14 at 20:36

AnasAnas

255

$endgroup$

add a comment | 

1

$begingroup$

Let $X$ be an exponential distribution of parameter $1$.

And let $Y$ be the binomial distribution of parameter $n$ and $p=1/2$.

We define $Z=frac{X}{Y+1}$. I'm trying to find the distribution of $Z$.

This is what I did :

Let $t>0$ we have $F(Z<t)=F(X<t(Y+1))=int_{0}^{t(Y+1)}exp(-x)dx=1-exp(-t(Y+1))$

and for $t<0$ we have $F(Z<t)=0.$

probability probability-theory random-variables

share|cite|improve this question

asked Mar 14 at 20:36

AnasAnas

255

$endgroup$

add a comment | 

$begingroup$

Let $X$ be an exponential distribution of parameter $1$.

And let $Y$ be the binomial distribution of parameter $n$ and $p=1/2$.

We define $Z=frac{X}{Y+1}$. I'm trying to find the distribution of $Z$.

This is what I did :

Let $t>0$ we have $F(Z<t)=F(X<t(Y+1))=int_{0}^{t(Y+1)}exp(-x)dx=1-exp(-t(Y+1))$

and for $t<0$ we have $F(Z<t)=0.$

probability probability-theory random-variables

share|cite|improve this question

asked Mar 14 at 20:36

AnasAnas

255

$endgroup$

share|cite|improve this question

asked Mar 14 at 20:36

AnasAnas

255

share|cite|improve this question

asked Mar 14 at 20:36

AnasAnas

255

asked Mar 14 at 20:36

AnasAnas

255

asked Mar 14 at 20:36

AnasAnas

255

1 Answer
1

active

oldest

votes

1

$begingroup$

You also need to sum over all possible values of Y.

Let's condition over $Y$. We get:

$$P(Z<t) = sumlimits_k P(Z<t|Y=k) P(Y=k)$$

So your net probability will become (using the fact that $Y$ is binomial):

$$P(Z<t) = sumlimits_{k=0}^n (1-e^{-t(k+1)}) {n choose k} p^k(1-p)^{n-k}$$

$$ = sumlimits_{k=0}^n {n choose k}p^k(1-p)^{n-k}- e^{-t}sumlimits_{k=0}^n {n choose k} (e^{-t}p)^k(1-p)^{n-k}$$
$$ = 1-e^{-t}sumlimits_{k=0}^n {n choose k} (e^{-t}p)^k(1-p)^{n-k}$$

share|cite|improve this answer

edited Mar 14 at 21:38

answered Mar 14 at 20:46

Rohit PandeyRohit Pandey

1,6331023

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did we sum over all possible values of $Y$, instead of sum from $k=0$ to for example $k=j$ with $j<n$ ?
  $endgroup$
  – Anas
  Mar 14 at 21:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Pandey can you please just write where exactly the first line come from ?
  $endgroup$
  – Anas
  Mar 14 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My counter question to that would be, why would we not sum over all possible values of $Y$? There is a probability that this value of $Y$ will be realized (whatever it might be) and conditional on that value of $Y$, there will be a certain probability $Z<t$.
  $endgroup$
  – Rohit Pandey
  Mar 14 at 21:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes makes sens, thank you
  $endgroup$
  – Anas
  Mar 14 at 21:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If this addressed your query, don't forget to mark as accepted answer :)
  $endgroup$
  – Rohit Pandey
  Mar 14 at 21:19
  

  
  
  
  

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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

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%2fmath.stackexchange.com%2fquestions%2f3148471%2fmixed-random-variable-exponential-distribution-mixed-with-binomial-distribution%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

You also need to sum over all possible values of Y.

Let's condition over $Y$. We get:

$$P(Z<t) = sumlimits_k P(Z<t|Y=k) P(Y=k)$$

So your net probability will become (using the fact that $Y$ is binomial):

$$P(Z<t) = sumlimits_{k=0}^n (1-e^{-t(k+1)}) {n choose k} p^k(1-p)^{n-k}$$

$$ = sumlimits_{k=0}^n {n choose k}p^k(1-p)^{n-k}- e^{-t}sumlimits_{k=0}^n {n choose k} (e^{-t}p)^k(1-p)^{n-k}$$
$$ = 1-e^{-t}sumlimits_{k=0}^n {n choose k} (e^{-t}p)^k(1-p)^{n-k}$$

share|cite|improve this answer

edited Mar 14 at 21:38

answered Mar 14 at 20:46

Rohit PandeyRohit Pandey

1,6331023

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did we sum over all possible values of $Y$, instead of sum from $k=0$ to for example $k=j$ with $j<n$ ?
  $endgroup$
  – Anas
  Mar 14 at 21:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Pandey can you please just write where exactly the first line come from ?
  $endgroup$
  – Anas
  Mar 14 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My counter question to that would be, why would we not sum over all possible values of $Y$? There is a probability that this value of $Y$ will be realized (whatever it might be) and conditional on that value of $Y$, there will be a certain probability $Z<t$.
  $endgroup$
  – Rohit Pandey
  Mar 14 at 21:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes makes sens, thank you
  $endgroup$
  – Anas
  Mar 14 at 21:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If this addressed your query, don't forget to mark as accepted answer :)
  $endgroup$
  – Rohit Pandey
  Mar 14 at 21:19
  

  
  
  
  

add a comment | 

1

$begingroup$

You also need to sum over all possible values of Y.

Let's condition over $Y$. We get:

$$P(Z<t) = sumlimits_k P(Z<t|Y=k) P(Y=k)$$

So your net probability will become (using the fact that $Y$ is binomial):

$$P(Z<t) = sumlimits_{k=0}^n (1-e^{-t(k+1)}) {n choose k} p^k(1-p)^{n-k}$$

$$ = sumlimits_{k=0}^n {n choose k}p^k(1-p)^{n-k}- e^{-t}sumlimits_{k=0}^n {n choose k} (e^{-t}p)^k(1-p)^{n-k}$$
$$ = 1-e^{-t}sumlimits_{k=0}^n {n choose k} (e^{-t}p)^k(1-p)^{n-k}$$

share|cite|improve this answer

edited Mar 14 at 21:38

answered Mar 14 at 20:46

Rohit PandeyRohit Pandey

1,6331023

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why did we sum over all possible values of $Y$, instead of sum from $k=0$ to for example $k=j$ with $j<n$ ?
  $endgroup$
  – Anas
  Mar 14 at 21:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Pandey can you please just write where exactly the first line come from ?
  $endgroup$
  – Anas
  Mar 14 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My counter question to that would be, why would we not sum over all possible values of $Y$? There is a probability that this value of $Y$ will be realized (whatever it might be) and conditional on that value of $Y$, there will be a certain probability $Z<t$.
  $endgroup$
  – Rohit Pandey
  Mar 14 at 21:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes makes sens, thank you
  $endgroup$
  – Anas
  Mar 14 at 21:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If this addressed your query, don't forget to mark as accepted answer :)
  $endgroup$
  – Rohit Pandey
  Mar 14 at 21:19
  

  
  
  
  

add a comment | 

$begingroup$

You also need to sum over all possible values of Y.

Let's condition over $Y$. We get:

$$P(Z<t) = sumlimits_k P(Z<t|Y=k) P(Y=k)$$

So your net probability will become (using the fact that $Y$ is binomial):

$$P(Z<t) = sumlimits_{k=0}^n (1-e^{-t(k+1)}) {n choose k} p^k(1-p)^{n-k}$$

$$ = sumlimits_{k=0}^n {n choose k}p^k(1-p)^{n-k}- e^{-t}sumlimits_{k=0}^n {n choose k} (e^{-t}p)^k(1-p)^{n-k}$$
$$ = 1-e^{-t}sumlimits_{k=0}^n {n choose k} (e^{-t}p)^k(1-p)^{n-k}$$

share|cite|improve this answer

edited Mar 14 at 21:38

answered Mar 14 at 20:46

Rohit PandeyRohit Pandey

1,6331023

$endgroup$

share|cite|improve this answer

edited Mar 14 at 21:38

answered Mar 14 at 20:46

Rohit PandeyRohit Pandey

1,6331023

answered Mar 14 at 20:46

Rohit PandeyRohit Pandey

1,6331023

answered Mar 14 at 20:46

Rohit PandeyRohit Pandey

1,6331023