Mixing
given $mathbb{P}(X=n, Y leq t)$ find $mathbb{P}(X=n)$.
$begingroup$
$X, Y$ are random variables
let $a,b > 0$, $ operatorname{supp}(X,Y) = mathbb{N} times mathbb{R}_{+}$ and
$$mathbb{P}(X=n, Y leq t) = int_0^{t}frac{(ay)^n}{n!}exp(-(a+b)y),dy$$
my approach consists of finding the joint CDF then taking the limit as $tto infty$ to obtain $X$'s CDF and then hoping to end up with a well known CDF, however it failed :
$$mathbb{P}(X leq n, Y leq t) =sum_{k = 0}^n mathbb{P}(X = k, Y leq t)$$
then $$begin{align}mathbb{P}(X leq n) &= lim_{ttoinfty} sum_{k = 0}^n mathbb{P}(X = k, Y leq t) \ &=lim_{ttoinfty}int_0^{t}sum_{k = 0}^n frac{(ay)^k}{k!}exp(-(a+b)y),dy \
&=int_0^{infty}sum_{k = 0}^n frac{(ay)^k}{k!}exp(-(a+b)y),dy \ &=sum_{k = 0}^nint_0^{infty} frac{(ay)^k}{k!}exp(-(a+b)y),dy \
&=sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}k!}int_0^{infty} z^{k}exp(-z),dz\
&=sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}}
end{align}$$
I think I'm doing something wrong because :
$$lim_{nto+infty} sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}} =frac{1}{b} neq 1$$
what am I doing wrong ? is it because we're dealing with a hybrid pair of random variables ?
probability-theory probability-distributions random-variables
edited Feb 2 at 15:31
Bernard
124k741117
asked Feb 2 at 15:27
rapidracimrapidracim
1,7541419
$endgroup$
add a comment |
$begingroup$
$X, Y$ are random variables
let $a,b > 0$, $ operatorname{supp}(X,Y) = mathbb{N} times mathbb{R}_{+}$ and
$$mathbb{P}(X=n, Y leq t) = int_0^{t}frac{(ay)^n}{n!}exp(-(a+b)y),dy$$
my approach consists of finding the joint CDF then taking the limit as $tto infty$ to obtain $X$'s CDF and then hoping to end up with a well known CDF, however it failed :
$$mathbb{P}(X leq n, Y leq t) =sum_{k = 0}^n mathbb{P}(X = k, Y leq t)$$
then $$begin{align}mathbb{P}(X leq n) &= lim_{ttoinfty} sum_{k = 0}^n mathbb{P}(X = k, Y leq t) \ &=lim_{ttoinfty}int_0^{t}sum_{k = 0}^n frac{(ay)^k}{k!}exp(-(a+b)y),dy \
&=int_0^{infty}sum_{k = 0}^n frac{(ay)^k}{k!}exp(-(a+b)y),dy \ &=sum_{k = 0}^nint_0^{infty} frac{(ay)^k}{k!}exp(-(a+b)y),dy \
&=sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}k!}int_0^{infty} z^{k}exp(-z),dz\
&=sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}}
end{align}$$
I think I'm doing something wrong because :
$$lim_{nto+infty} sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}} =frac{1}{b} neq 1$$
what am I doing wrong ? is it because we're dealing with a hybrid pair of random variables ?
probability-theory probability-distributions random-variables
edited Feb 2 at 15:31
Bernard
124k741117
asked Feb 2 at 15:27
rapidracimrapidracim
1,7541419
$endgroup$
$begingroup$
?? Why the détour? $$P(X=n)=lim_{ttoinfty}P(X=n,Yleqslant t)=ldots$$ the trouble being that the formula for $P(X=n,Yleqslant t)$ you are given/giving is wrong.
$endgroup$
– Did
Feb 2 at 15:34
$begingroup$
@Did I did that at first but ended up with $mathbb{P}(x = n) = frac{a^n}{(a+b)^{n+1}}$ so same issue
$endgroup$
– rapidracim
Feb 2 at 15:36
1
$begingroup$
The correct formula might be $$P(X=n, Y leqslant t) = int_0^tbfrac{(ay)^n}{n!}e^{-(a+b)y},dy$$
$endgroup$
– Did
Feb 2 at 15:37
$begingroup$
...Anyway, passing by the CDF was not going to make the problem disappear.
$endgroup$
– Did
Feb 2 at 15:42
$begingroup$
@Did yes I have to admit the détour was stupid, anyways I checked again and the problem was written exactly like this by my teacher so it means he did a typo. now the problem being ill-stated, should I delete the question ?
$endgroup$
– rapidracim
Feb 2 at 15:45
add a comment |
$begingroup$
$X, Y$ are random variables
let $a,b > 0$, $ operatorname{supp}(X,Y) = mathbb{N} times mathbb{R}_{+}$ and
$$mathbb{P}(X=n, Y leq t) = int_0^{t}frac{(ay)^n}{n!}exp(-(a+b)y),dy$$
my approach consists of finding the joint CDF then taking the limit as $tto infty$ to obtain $X$'s CDF and then hoping to end up with a well known CDF, however it failed :
$$mathbb{P}(X leq n, Y leq t) =sum_{k = 0}^n mathbb{P}(X = k, Y leq t)$$
then $$begin{align}mathbb{P}(X leq n) &= lim_{ttoinfty} sum_{k = 0}^n mathbb{P}(X = k, Y leq t) \ &=lim_{ttoinfty}int_0^{t}sum_{k = 0}^n frac{(ay)^k}{k!}exp(-(a+b)y),dy \
&=int_0^{infty}sum_{k = 0}^n frac{(ay)^k}{k!}exp(-(a+b)y),dy \ &=sum_{k = 0}^nint_0^{infty} frac{(ay)^k}{k!}exp(-(a+b)y),dy \
&=sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}k!}int_0^{infty} z^{k}exp(-z),dz\
&=sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}}
end{align}$$
I think I'm doing something wrong because :
$$lim_{nto+infty} sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}} =frac{1}{b} neq 1$$
what am I doing wrong ? is it because we're dealing with a hybrid pair of random variables ?
probability-theory probability-distributions random-variables
edited Feb 2 at 15:31
Bernard
124k741117
asked Feb 2 at 15:27
rapidracimrapidracim
1,7541419
$endgroup$
$X, Y$ are random variables
let $a,b > 0$, $ operatorname{supp}(X,Y) = mathbb{N} times mathbb{R}_{+}$ and
$$mathbb{P}(X=n, Y leq t) = int_0^{t}frac{(ay)^n}{n!}exp(-(a+b)y),dy$$
my approach consists of finding the joint CDF then taking the limit as $tto infty$ to obtain $X$'s CDF and then hoping to end up with a well known CDF, however it failed :
$$mathbb{P}(X leq n, Y leq t) =sum_{k = 0}^n mathbb{P}(X = k, Y leq t)$$
then $$begin{align}mathbb{P}(X leq n) &= lim_{ttoinfty} sum_{k = 0}^n mathbb{P}(X = k, Y leq t) \ &=lim_{ttoinfty}int_0^{t}sum_{k = 0}^n frac{(ay)^k}{k!}exp(-(a+b)y),dy \
&=int_0^{infty}sum_{k = 0}^n frac{(ay)^k}{k!}exp(-(a+b)y),dy \ &=sum_{k = 0}^nint_0^{infty} frac{(ay)^k}{k!}exp(-(a+b)y),dy \
&=sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}k!}int_0^{infty} z^{k}exp(-z),dz\
&=sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}}
end{align}$$
I think I'm doing something wrong because :
$$lim_{nto+infty} sum_{k = 0}^n frac{a^k}{(a+b)^{k+1}} =frac{1}{b} neq 1$$
what am I doing wrong ? is it because we're dealing with a hybrid pair of random variables ?
probability-theory probability-distributions random-variables
probability-theory probability-distributions random-variables
edited Feb 2 at 15:31
Bernard
124k741117
asked Feb 2 at 15:27
rapidracimrapidracim
1,7541419
edited Feb 2 at 15:31
Bernard
124k741117
asked Feb 2 at 15:27
rapidracimrapidracim
1,7541419
edited Feb 2 at 15:31
Bernard
124k741117
edited Feb 2 at 15:31
Bernard
124k741117
edited Feb 2 at 15:31
Bernard
124k741117
124k741117
asked Feb 2 at 15:27
rapidracimrapidracim
1,7541419
asked Feb 2 at 15:27
rapidracimrapidracim
1,7541419
asked Feb 2 at 15:27
rapidracimrapidracim
1,7541419
1,7541419
$begingroup$
?? Why the détour? $$P(X=n)=lim_{ttoinfty}P(X=n,Yleqslant t)=ldots$$ the trouble being that the formula for $P(X=n,Yleqslant t)$ you are given/giving is wrong.
$endgroup$
– Did
Feb 2 at 15:34
$begingroup$
@Did I did that at first but ended up with $mathbb{P}(x = n) = frac{a^n}{(a+b)^{n+1}}$ so same issue
$endgroup$
– rapidracim
Feb 2 at 15:36
1
$begingroup$
The correct formula might be $$P(X=n, Y leqslant t) = int_0^tbfrac{(ay)^n}{n!}e^{-(a+b)y},dy$$
$endgroup$
– Did
Feb 2 at 15:37
$begingroup$
...Anyway, passing by the CDF was not going to make the problem disappear.
$endgroup$
– Did
Feb 2 at 15:42
$begingroup$
@Did yes I have to admit the détour was stupid, anyways I checked again and the problem was written exactly like this by my teacher so it means he did a typo. now the problem being ill-stated, should I delete the question ?
$endgroup$
– rapidracim
Feb 2 at 15:45
add a comment |
$begingroup$
?? Why the détour? $$P(X=n)=lim_{ttoinfty}P(X=n,Yleqslant t)=ldots$$ the trouble being that the formula for $P(X=n,Yleqslant t)$ you are given/giving is wrong.
$endgroup$
– Did
Feb 2 at 15:34
$begingroup$
@Did I did that at first but ended up with $mathbb{P}(x = n) = frac{a^n}{(a+b)^{n+1}}$ so same issue
$endgroup$
– rapidracim
Feb 2 at 15:36
1
$begingroup$
The correct formula might be $$P(X=n, Y leqslant t) = int_0^tbfrac{(ay)^n}{n!}e^{-(a+b)y},dy$$
$endgroup$
– Did
Feb 2 at 15:37
$begingroup$
...Anyway, passing by the CDF was not going to make the problem disappear.
$endgroup$
– Did
Feb 2 at 15:42
$begingroup$
@Did yes I have to admit the détour was stupid, anyways I checked again and the problem was written exactly like this by my teacher so it means he did a typo. now the problem being ill-stated, should I delete the question ?
$endgroup$
– rapidracim
Feb 2 at 15:45
$begingroup$
?? Why the détour? $$P(X=n)=lim_{ttoinfty}P(X=n,Yleqslant t)=ldots$$ the trouble being that the formula for $P(X=n,Yleqslant t)$ you are given/giving is wrong.
$endgroup$
– Did
Feb 2 at 15:34
$begingroup$
?? Why the détour? $$P(X=n)=lim_{ttoinfty}P(X=n,Yleqslant t)=ldots$$ the trouble being that the formula for $P(X=n,Yleqslant t)$ you are given/giving is wrong.
$endgroup$
– Did
Feb 2 at 15:34
$begingroup$
@Did I did that at first but ended up with $mathbb{P}(x = n) = frac{a^n}{(a+b)^{n+1}}$ so same issue
$endgroup$
– rapidracim
Feb 2 at 15:36
$begingroup$
@Did I did that at first but ended up with $mathbb{P}(x = n) = frac{a^n}{(a+b)^{n+1}}$ so same issue
$endgroup$
– rapidracim
Feb 2 at 15:36
1
1
$begingroup$
The correct formula might be $$P(X=n, Y leqslant t) = int_0^tbfrac{(ay)^n}{n!}e^{-(a+b)y},dy$$
$endgroup$
– Did
Feb 2 at 15:37
$begingroup$
The correct formula might be $$P(X=n, Y leqslant t) = int_0^tbfrac{(ay)^n}{n!}e^{-(a+b)y},dy$$
$endgroup$
– Did
Feb 2 at 15:37
$begingroup$
...Anyway, passing by the CDF was not going to make the problem disappear.
$endgroup$
– Did
Feb 2 at 15:42
$begingroup$
...Anyway, passing by the CDF was not going to make the problem disappear.
$endgroup$
– Did
Feb 2 at 15:42
$begingroup$
@Did yes I have to admit the détour was stupid, anyways I checked again and the problem was written exactly like this by my teacher so it means he did a typo. now the problem being ill-stated, should I delete the question ?
$endgroup$
– rapidracim
Feb 2 at 15:45
$begingroup$
@Did yes I have to admit the détour was stupid, anyways I checked again and the problem was written exactly like this by my teacher so it means he did a typo. now the problem being ill-stated, should I delete the question ?
$endgroup$
– rapidracim
Feb 2 at 15:45
add a comment |
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$begingroup$
?? Why the détour? $$P(X=n)=lim_{ttoinfty}P(X=n,Yleqslant t)=ldots$$ the trouble being that the formula for $P(X=n,Yleqslant t)$ you are given/giving is wrong.
$endgroup$
– Did
Feb 2 at 15:34
$begingroup$
@Did I did that at first but ended up with $mathbb{P}(x = n) = frac{a^n}{(a+b)^{n+1}}$ so same issue
$endgroup$
– rapidracim
Feb 2 at 15:36
1
$begingroup$
The correct formula might be $$P(X=n, Y leqslant t) = int_0^tbfrac{(ay)^n}{n!}e^{-(a+b)y},dy$$
$endgroup$
– Did
Feb 2 at 15:37
$begingroup$
...Anyway, passing by the CDF was not going to make the problem disappear.
$endgroup$
– Did
Feb 2 at 15:42
$begingroup$
@Did yes I have to admit the détour was stupid, anyways I checked again and the problem was written exactly like this by my teacher so it means he did a typo. now the problem being ill-stated, should I delete the question ?
$endgroup$
– rapidracim
Feb 2 at 15:45