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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 | 

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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