Mixing
Conditional distribution of the sum of two independent binomial variables [closed]
$begingroup$
I have a doubt about the following exercise:
Considering a Y, equal to the sum of two binomials (X1 and X2) with the same parameter p, what would be the conditional distribution of X1 given Y, in particular considering the case of Y = 3?
Thanks in advance for the help.
probability probability-theory discrete-mathematics
asked Jan 23 at 12:57
Umberto ManzaliniUmberto Manzalini
6
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closed as off-topic by Namaste, Lee David Chung Lin, metamorphy, Gibbs, Trevor Gunn Jan 23 at 22:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Lee David Chung Lin, metamorphy, Gibbs, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have a doubt about the following exercise:
Considering a Y, equal to the sum of two binomials (X1 and X2) with the same parameter p, what would be the conditional distribution of X1 given Y, in particular considering the case of Y = 3?
Thanks in advance for the help.
probability probability-theory discrete-mathematics
asked Jan 23 at 12:57
Umberto ManzaliniUmberto Manzalini
6
$endgroup$
closed as off-topic by Namaste, Lee David Chung Lin, metamorphy, Gibbs, Trevor Gunn Jan 23 at 22:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Lee David Chung Lin, metamorphy, Gibbs, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have a doubt about the following exercise:
Considering a Y, equal to the sum of two binomials (X1 and X2) with the same parameter p, what would be the conditional distribution of X1 given Y, in particular considering the case of Y = 3?
Thanks in advance for the help.
probability probability-theory discrete-mathematics
asked Jan 23 at 12:57
Umberto ManzaliniUmberto Manzalini
6
$endgroup$
I have a doubt about the following exercise:
Considering a Y, equal to the sum of two binomials (X1 and X2) with the same parameter p, what would be the conditional distribution of X1 given Y, in particular considering the case of Y = 3?
Thanks in advance for the help.
probability probability-theory discrete-mathematics
probability probability-theory discrete-mathematics
asked Jan 23 at 12:57
Umberto ManzaliniUmberto Manzalini
6
asked Jan 23 at 12:57
Umberto ManzaliniUmberto Manzalini
6
asked Jan 23 at 12:57
Umberto ManzaliniUmberto Manzalini
6
asked Jan 23 at 12:57
Umberto ManzaliniUmberto Manzalini
6
asked Jan 23 at 12:57
Umberto ManzaliniUmberto Manzalini
6
6
closed as off-topic by Namaste, Lee David Chung Lin, metamorphy, Gibbs, Trevor Gunn Jan 23 at 22:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Lee David Chung Lin, metamorphy, Gibbs, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Lee David Chung Lin, metamorphy, Gibbs, Trevor Gunn Jan 23 at 22:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Lee David Chung Lin, metamorphy, Gibbs, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Presumably you assume that $X_1$ and $X_2$ are independent?
That being the case: note that $P(X_1=x_1mid Y=3)=0$ if $x_1>3$. On the other hand, for $x_1leq 3$, start by writing
$$
P(X_1=x_1mid Y=3)=frac{P(X_1=x_1,Y=3)}{P(Y=3)}=frac{P(X_1=x_1,X_2=3-x_1)}{P(Y=3)}=frac{P(X_1=x_1)cdot P(X_2=3-x_1)}{P(Y=3)}.
$$
The numerator can be computed easily for each valid $x_1in{0,1,2,3}$. For the denominator, note that the sum of independent binomials $text{Bin}(n,p)$ and $text{Bin}(m,p)$ with the same probability is distributed as $text{Bin}(n+m,p)$.
answered Jan 23 at 16:10
Nick PetersonNick Peterson
26.8k23962
$endgroup$
$begingroup$
Thank you very much, they are independent; you are right. So I guess the support is RX1|Y= 0 <X1 <3-x2? But then the density function that derives from it would be that of a hypergeometric? because I am shown the result as such
$endgroup$
– Umberto Manzalini
Jan 23 at 16:35
$begingroup$
The support is $0leq X_1mid_{Y=3}leq 3$. You can easily write out the probabilities for each of these 4 possible values. :-)
$endgroup$
– Nick Peterson
Jan 23 at 16:52
$begingroup$
Thank you, but I am asked to recognize the distribution too? How should i do?
$endgroup$
– Umberto Manzalini
Jan 23 at 16:55
$begingroup$
Step 1: find the probabilities.
$endgroup$
– Nick Peterson
Jan 23 at 16:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Presumably you assume that $X_1$ and $X_2$ are independent?
That being the case: note that $P(X_1=x_1mid Y=3)=0$ if $x_1>3$. On the other hand, for $x_1leq 3$, start by writing
$$
P(X_1=x_1mid Y=3)=frac{P(X_1=x_1,Y=3)}{P(Y=3)}=frac{P(X_1=x_1,X_2=3-x_1)}{P(Y=3)}=frac{P(X_1=x_1)cdot P(X_2=3-x_1)}{P(Y=3)}.
$$
The numerator can be computed easily for each valid $x_1in{0,1,2,3}$. For the denominator, note that the sum of independent binomials $text{Bin}(n,p)$ and $text{Bin}(m,p)$ with the same probability is distributed as $text{Bin}(n+m,p)$.
answered Jan 23 at 16:10
Nick PetersonNick Peterson
26.8k23962
$endgroup$
$begingroup$
Thank you very much, they are independent; you are right. So I guess the support is RX1|Y= 0 <X1 <3-x2? But then the density function that derives from it would be that of a hypergeometric? because I am shown the result as such
$endgroup$
– Umberto Manzalini
Jan 23 at 16:35
$begingroup$
The support is $0leq X_1mid_{Y=3}leq 3$. You can easily write out the probabilities for each of these 4 possible values. :-)
$endgroup$
– Nick Peterson
Jan 23 at 16:52
$begingroup$
Thank you, but I am asked to recognize the distribution too? How should i do?
$endgroup$
– Umberto Manzalini
Jan 23 at 16:55
$begingroup$
Step 1: find the probabilities.
$endgroup$
– Nick Peterson
Jan 23 at 16:56
add a comment |
$begingroup$
Presumably you assume that $X_1$ and $X_2$ are independent?
That being the case: note that $P(X_1=x_1mid Y=3)=0$ if $x_1>3$. On the other hand, for $x_1leq 3$, start by writing
$$
P(X_1=x_1mid Y=3)=frac{P(X_1=x_1,Y=3)}{P(Y=3)}=frac{P(X_1=x_1,X_2=3-x_1)}{P(Y=3)}=frac{P(X_1=x_1)cdot P(X_2=3-x_1)}{P(Y=3)}.
$$
The numerator can be computed easily for each valid $x_1in{0,1,2,3}$. For the denominator, note that the sum of independent binomials $text{Bin}(n,p)$ and $text{Bin}(m,p)$ with the same probability is distributed as $text{Bin}(n+m,p)$.
answered Jan 23 at 16:10
Nick PetersonNick Peterson
26.8k23962
$endgroup$
$begingroup$
Thank you very much, they are independent; you are right. So I guess the support is RX1|Y= 0 <X1 <3-x2? But then the density function that derives from it would be that of a hypergeometric? because I am shown the result as such
$endgroup$
– Umberto Manzalini
Jan 23 at 16:35
$begingroup$
The support is $0leq X_1mid_{Y=3}leq 3$. You can easily write out the probabilities for each of these 4 possible values. :-)
$endgroup$
– Nick Peterson
Jan 23 at 16:52
$begingroup$
Thank you, but I am asked to recognize the distribution too? How should i do?
$endgroup$
– Umberto Manzalini
Jan 23 at 16:55
$begingroup$
Step 1: find the probabilities.
$endgroup$
– Nick Peterson
Jan 23 at 16:56
add a comment |
$begingroup$
Presumably you assume that $X_1$ and $X_2$ are independent?
That being the case: note that $P(X_1=x_1mid Y=3)=0$ if $x_1>3$. On the other hand, for $x_1leq 3$, start by writing
$$
P(X_1=x_1mid Y=3)=frac{P(X_1=x_1,Y=3)}{P(Y=3)}=frac{P(X_1=x_1,X_2=3-x_1)}{P(Y=3)}=frac{P(X_1=x_1)cdot P(X_2=3-x_1)}{P(Y=3)}.
$$
The numerator can be computed easily for each valid $x_1in{0,1,2,3}$. For the denominator, note that the sum of independent binomials $text{Bin}(n,p)$ and $text{Bin}(m,p)$ with the same probability is distributed as $text{Bin}(n+m,p)$.
answered Jan 23 at 16:10
Nick PetersonNick Peterson
26.8k23962
$endgroup$
Presumably you assume that $X_1$ and $X_2$ are independent?
That being the case: note that $P(X_1=x_1mid Y=3)=0$ if $x_1>3$. On the other hand, for $x_1leq 3$, start by writing
$$
P(X_1=x_1mid Y=3)=frac{P(X_1=x_1,Y=3)}{P(Y=3)}=frac{P(X_1=x_1,X_2=3-x_1)}{P(Y=3)}=frac{P(X_1=x_1)cdot P(X_2=3-x_1)}{P(Y=3)}.
$$
The numerator can be computed easily for each valid $x_1in{0,1,2,3}$. For the denominator, note that the sum of independent binomials $text{Bin}(n,p)$ and $text{Bin}(m,p)$ with the same probability is distributed as $text{Bin}(n+m,p)$.
answered Jan 23 at 16:10
Nick PetersonNick Peterson
26.8k23962
answered Jan 23 at 16:10
Nick PetersonNick Peterson
26.8k23962
answered Jan 23 at 16:10
Nick PetersonNick Peterson
26.8k23962
answered Jan 23 at 16:10
Nick PetersonNick Peterson
26.8k23962
26.8k23962
$begingroup$
Thank you very much, they are independent; you are right. So I guess the support is RX1|Y= 0 <X1 <3-x2? But then the density function that derives from it would be that of a hypergeometric? because I am shown the result as such
$endgroup$
– Umberto Manzalini
Jan 23 at 16:35
$begingroup$
The support is $0leq X_1mid_{Y=3}leq 3$. You can easily write out the probabilities for each of these 4 possible values. :-)
$endgroup$
– Nick Peterson
Jan 23 at 16:52
$begingroup$
Thank you, but I am asked to recognize the distribution too? How should i do?
$endgroup$
– Umberto Manzalini
Jan 23 at 16:55
$begingroup$
Step 1: find the probabilities.
$endgroup$
– Nick Peterson
Jan 23 at 16:56
add a comment |
$begingroup$
Thank you very much, they are independent; you are right. So I guess the support is RX1|Y= 0 <X1 <3-x2? But then the density function that derives from it would be that of a hypergeometric? because I am shown the result as such
$endgroup$
– Umberto Manzalini
Jan 23 at 16:35
$begingroup$
The support is $0leq X_1mid_{Y=3}leq 3$. You can easily write out the probabilities for each of these 4 possible values. :-)
$endgroup$
– Nick Peterson
Jan 23 at 16:52
$begingroup$
Thank you, but I am asked to recognize the distribution too? How should i do?
$endgroup$
– Umberto Manzalini
Jan 23 at 16:55
$begingroup$
Step 1: find the probabilities.
$endgroup$
– Nick Peterson
Jan 23 at 16:56
$begingroup$
Thank you very much, they are independent; you are right. So I guess the support is RX1|Y= 0 <X1 <3-x2? But then the density function that derives from it would be that of a hypergeometric? because I am shown the result as such
$endgroup$
– Umberto Manzalini
Jan 23 at 16:35
$begingroup$
Thank you very much, they are independent; you are right. So I guess the support is RX1|Y= 0 <X1 <3-x2? But then the density function that derives from it would be that of a hypergeometric? because I am shown the result as such
$endgroup$
– Umberto Manzalini
Jan 23 at 16:35
$begingroup$
The support is $0leq X_1mid_{Y=3}leq 3$. You can easily write out the probabilities for each of these 4 possible values. :-)
$endgroup$
– Nick Peterson
Jan 23 at 16:52
$begingroup$
The support is $0leq X_1mid_{Y=3}leq 3$. You can easily write out the probabilities for each of these 4 possible values. :-)
$endgroup$
– Nick Peterson
Jan 23 at 16:52
$begingroup$
Thank you, but I am asked to recognize the distribution too? How should i do?
$endgroup$
– Umberto Manzalini
Jan 23 at 16:55
$begingroup$
Thank you, but I am asked to recognize the distribution too? How should i do?
$endgroup$
– Umberto Manzalini
Jan 23 at 16:55
$begingroup$
Step 1: find the probabilities.
$endgroup$
– Nick Peterson
Jan 23 at 16:56
$begingroup$
Step 1: find the probabilities.
$endgroup$
– Nick Peterson
Jan 23 at 16:56
add a comment |
