Find $P(X=Y+1)$
up vote
1
down vote
favorite
Suppose that X and Y are integer valued random variables with joint probability mass function given by
$p_{X,Y}(a, b)=begin{cases} frac{1}{4a}, & 1leq bleq aleq 4\ 0, & text{otherwise}. end{cases}.$
(b) Find the marginal probability mass functions of X and Y.
(c) Find $P(X=Y+1)$
Since this is a discrete random variable, I want to construct a table of joint distribution to find the sum of rows, columns for pmf of X and Y, but I'm having trouble with it. Anyone can help with part (b) and (c)?
probability
asked Nov 14 at 4:18
dxdydz
989
add a comment |
up vote
1
down vote
favorite
Suppose that X and Y are integer valued random variables with joint probability mass function given by
$p_{X,Y}(a, b)=begin{cases} frac{1}{4a}, & 1leq bleq aleq 4\ 0, & text{otherwise}. end{cases}.$
(b) Find the marginal probability mass functions of X and Y.
(c) Find $P(X=Y+1)$
Since this is a discrete random variable, I want to construct a table of joint distribution to find the sum of rows, columns for pmf of X and Y, but I'm having trouble with it. Anyone can help with part (b) and (c)?
probability
asked Nov 14 at 4:18
dxdydz
989
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that X and Y are integer valued random variables with joint probability mass function given by
$p_{X,Y}(a, b)=begin{cases} frac{1}{4a}, & 1leq bleq aleq 4\ 0, & text{otherwise}. end{cases}.$
(b) Find the marginal probability mass functions of X and Y.
(c) Find $P(X=Y+1)$
Since this is a discrete random variable, I want to construct a table of joint distribution to find the sum of rows, columns for pmf of X and Y, but I'm having trouble with it. Anyone can help with part (b) and (c)?
probability
asked Nov 14 at 4:18
dxdydz
989
Suppose that X and Y are integer valued random variables with joint probability mass function given by
$p_{X,Y}(a, b)=begin{cases} frac{1}{4a}, & 1leq bleq aleq 4\ 0, & text{otherwise}. end{cases}.$
(b) Find the marginal probability mass functions of X and Y.
(c) Find $P(X=Y+1)$
Since this is a discrete random variable, I want to construct a table of joint distribution to find the sum of rows, columns for pmf of X and Y, but I'm having trouble with it. Anyone can help with part (b) and (c)?
probability
probability
asked Nov 14 at 4:18
dxdydz
989
asked Nov 14 at 4:18
dxdydz
989
asked Nov 14 at 4:18
dxdydz
989
asked Nov 14 at 4:18
dxdydz
989
asked Nov 14 at 4:18
dxdydz
989
989
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The table is not hard. Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .
$$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$
answered Nov 14 at 5:00
Graham Kemp
84.2k43378
1
So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
– dxdydz
Nov 14 at 5:08
Indeed. @dxdydz
– Graham Kemp
Nov 14 at 11:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The table is not hard. Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .
$$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$
answered Nov 14 at 5:00
Graham Kemp
84.2k43378
1
So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
– dxdydz
Nov 14 at 5:08
Indeed. @dxdydz
– Graham Kemp
Nov 14 at 11:59
add a comment |
up vote
1
down vote
accepted
The table is not hard. Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .
$$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$
answered Nov 14 at 5:00
Graham Kemp
84.2k43378
1
So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
– dxdydz
Nov 14 at 5:08
Indeed. @dxdydz
– Graham Kemp
Nov 14 at 11:59
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The table is not hard. Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .
$$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$
answered Nov 14 at 5:00
Graham Kemp
84.2k43378
The table is not hard. Each cell in 4 rows ($1leq aleq 4$) of $a$ columns ($1leq bleq a$), contains $tfrac 1{4a}$, the rest of the cells are zero .
$$p_{X,Y}(a,b)=tfrac 1{4a}mathbf 1_{1leq bleq aleq 4}\boxed{begin{array}{c|c:c:c:c|c}abackslash b & 1 & 2 & 3 & 4 &\hline 1 & tfrac 14 &0&0&0\ hdashline 2 & tfrac 18&tfrac 18&0&0\ hdashline 3 &tfrac 1{12}&tfrac 1{12}&tfrac 1{12}&0\ hdashline 4 & tfrac 1{16}&tfrac 1{16}&tfrac 1{16}&tfrac 1{16}\hline ~&&&&&1end{array}}$$
answered Nov 14 at 5:00
Graham Kemp
84.2k43378
edited Nov 14 at 5:06
answered Nov 14 at 5:00
Graham Kemp
84.2k43378
answered Nov 14 at 5:00
Graham Kemp
84.2k43378
answered Nov 14 at 5:00
Graham Kemp
84.2k43378
84.2k43378
1
So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
– dxdydz
Nov 14 at 5:08
Indeed. @dxdydz
– Graham Kemp
Nov 14 at 11:59
add a comment |
1
So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
– dxdydz
Nov 14 at 5:08
Indeed. @dxdydz
– Graham Kemp
Nov 14 at 11:59
1
1
So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
– dxdydz
Nov 14 at 5:08
So $P(X=Y+1) = P(X=2, Y=1) + P(X=3, Y=2) + P(X=4, Y=3) = 1/8 + 1/12 +1/16 = 13/48$?
– dxdydz
Nov 14 at 5:08
Indeed. @dxdydz
– Graham Kemp
Nov 14 at 11:59
Indeed. @dxdydz
– Graham Kemp
Nov 14 at 11:59
add a comment |
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997775%2ffind-px-y1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
