Transition Probabilities - Markov Process [closed]
$begingroup$
Suppose there is a box with $N$ balls. Each ball is coloured either red or blue. In each time period, one ball is chosen at random from the box and with probability $frac{1}{2}$ replaced with the other colour; or the ball is returned to the box. Let $X_{n}$ denote the number of red balls after $n$ picks.
A) Determine the transition probabilities: $$p_{ij}, forall{i,j}=0,1,....,N$$
Any help would be much appreciated thank you.
probability probability-theory stochastic-processes markov-chains markov-process
asked Jan 6 at 13:08
ReetyReety
13911
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closed as off-topic by Did, RRL, Holo, Martin R, Saad Jan 9 at 14:12
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." – Holo, Martin R, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose there is a box with $N$ balls. Each ball is coloured either red or blue. In each time period, one ball is chosen at random from the box and with probability $frac{1}{2}$ replaced with the other colour; or the ball is returned to the box. Let $X_{n}$ denote the number of red balls after $n$ picks.
A) Determine the transition probabilities: $$p_{ij}, forall{i,j}=0,1,....,N$$
Any help would be much appreciated thank you.
probability probability-theory stochastic-processes markov-chains markov-process
asked Jan 6 at 13:08
ReetyReety
13911
$endgroup$
closed as off-topic by Did, RRL, Holo, Martin R, Saad Jan 9 at 14:12
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." – Holo, Martin R, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose there is a box with $N$ balls. Each ball is coloured either red or blue. In each time period, one ball is chosen at random from the box and with probability $frac{1}{2}$ replaced with the other colour; or the ball is returned to the box. Let $X_{n}$ denote the number of red balls after $n$ picks.
A) Determine the transition probabilities: $$p_{ij}, forall{i,j}=0,1,....,N$$
Any help would be much appreciated thank you.
probability probability-theory stochastic-processes markov-chains markov-process
asked Jan 6 at 13:08
ReetyReety
13911
$endgroup$
Suppose there is a box with $N$ balls. Each ball is coloured either red or blue. In each time period, one ball is chosen at random from the box and with probability $frac{1}{2}$ replaced with the other colour; or the ball is returned to the box. Let $X_{n}$ denote the number of red balls after $n$ picks.
A) Determine the transition probabilities: $$p_{ij}, forall{i,j}=0,1,....,N$$
Any help would be much appreciated thank you.
probability probability-theory stochastic-processes markov-chains markov-process
probability probability-theory stochastic-processes markov-chains markov-process
asked Jan 6 at 13:08
ReetyReety
13911
asked Jan 6 at 13:08
ReetyReety
13911
edited Jan 6 at 14:53
Reety
asked Jan 6 at 13:08
ReetyReety
13911
asked Jan 6 at 13:08
ReetyReety
13911
asked Jan 6 at 13:08
ReetyReety
13911
13911
closed as off-topic by Did, RRL, Holo, Martin R, Saad Jan 9 at 14:12
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." – Holo, Martin R, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, RRL, Holo, Martin R, Saad Jan 9 at 14:12
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." – Holo, Martin R, Saad
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
$begingroup$
Hints:
If you have $x$ red balls after $n$ picks, what is the probability that your $(n+1)$th pick is red?
or blue?
What happens to the number of red balls in each of these cases?
If you have $x$ red balls after $n$ picks, what are the possible numbers of red balls after $n+1$ picks?
What are the probabilities that you transition to each of these possibilities?
answered Jan 6 at 15:19
HenryHenry
98.8k476163
$endgroup$
$begingroup$
okay so far for 1 i got $frac{x}{N}$ and 2 i got $frac{N-x}{N}$ and I have no idea what next.
$endgroup$
– Reety
Jan 6 at 16:19
1
$begingroup$
@Reety You may be overthinking this as you have already done the heavy work. For example with question 3 the number of red balls goes down by $1$ if the $(n+1)$th pick was red, and up by $1$ if the $(n+1)$th pick was blue. So for 4 the number of red balls becomes $x-1$ or $x+1$, and for 5 the respective probabilities of these are those you identified for 1 and 2; these are the transition probabilities as all others are $0$
$endgroup$
– Henry
Jan 6 at 18:17
$begingroup$
$p_{ii}=frac{1}{2}$, $p_{i(i-1)}=frac{x}{N}$, $p_{i(i+1)}=frac{N-x}{N}$ and $p_{ij}=0$ if |i-j|≥2 this is what I got
$endgroup$
– Reety
Jan 6 at 20:42
1
$begingroup$
@Reety I think you should have $p_{ii}=0$ but otherwise that seems correct. Perhaps better to write something like $p_{i,(i-1)}=frac{i}{N}, p_{i,(i+1)}=frac{N-i}{N}, p_{i,j}=0$ if $j not=ipm1$ as $x$ and $i$ represent the same thing
$endgroup$
– Henry
Jan 6 at 21:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hints:
If you have $x$ red balls after $n$ picks, what is the probability that your $(n+1)$th pick is red?
or blue?
What happens to the number of red balls in each of these cases?
If you have $x$ red balls after $n$ picks, what are the possible numbers of red balls after $n+1$ picks?
What are the probabilities that you transition to each of these possibilities?
answered Jan 6 at 15:19
HenryHenry
98.8k476163
$endgroup$
$begingroup$
okay so far for 1 i got $frac{x}{N}$ and 2 i got $frac{N-x}{N}$ and I have no idea what next.
$endgroup$
– Reety
Jan 6 at 16:19
1
$begingroup$
@Reety You may be overthinking this as you have already done the heavy work. For example with question 3 the number of red balls goes down by $1$ if the $(n+1)$th pick was red, and up by $1$ if the $(n+1)$th pick was blue. So for 4 the number of red balls becomes $x-1$ or $x+1$, and for 5 the respective probabilities of these are those you identified for 1 and 2; these are the transition probabilities as all others are $0$
$endgroup$
– Henry
Jan 6 at 18:17
$begingroup$
$p_{ii}=frac{1}{2}$, $p_{i(i-1)}=frac{x}{N}$, $p_{i(i+1)}=frac{N-x}{N}$ and $p_{ij}=0$ if |i-j|≥2 this is what I got
$endgroup$
– Reety
Jan 6 at 20:42
1
$begingroup$
@Reety I think you should have $p_{ii}=0$ but otherwise that seems correct. Perhaps better to write something like $p_{i,(i-1)}=frac{i}{N}, p_{i,(i+1)}=frac{N-i}{N}, p_{i,j}=0$ if $j not=ipm1$ as $x$ and $i$ represent the same thing
$endgroup$
– Henry
Jan 6 at 21:22
add a comment |
$begingroup$
Hints:
If you have $x$ red balls after $n$ picks, what is the probability that your $(n+1)$th pick is red?
or blue?
What happens to the number of red balls in each of these cases?
If you have $x$ red balls after $n$ picks, what are the possible numbers of red balls after $n+1$ picks?
What are the probabilities that you transition to each of these possibilities?
answered Jan 6 at 15:19
HenryHenry
98.8k476163
$endgroup$
$begingroup$
okay so far for 1 i got $frac{x}{N}$ and 2 i got $frac{N-x}{N}$ and I have no idea what next.
$endgroup$
– Reety
Jan 6 at 16:19
1
$begingroup$
@Reety You may be overthinking this as you have already done the heavy work. For example with question 3 the number of red balls goes down by $1$ if the $(n+1)$th pick was red, and up by $1$ if the $(n+1)$th pick was blue. So for 4 the number of red balls becomes $x-1$ or $x+1$, and for 5 the respective probabilities of these are those you identified for 1 and 2; these are the transition probabilities as all others are $0$
$endgroup$
– Henry
Jan 6 at 18:17
$begingroup$
$p_{ii}=frac{1}{2}$, $p_{i(i-1)}=frac{x}{N}$, $p_{i(i+1)}=frac{N-x}{N}$ and $p_{ij}=0$ if |i-j|≥2 this is what I got
$endgroup$
– Reety
Jan 6 at 20:42
1
$begingroup$
@Reety I think you should have $p_{ii}=0$ but otherwise that seems correct. Perhaps better to write something like $p_{i,(i-1)}=frac{i}{N}, p_{i,(i+1)}=frac{N-i}{N}, p_{i,j}=0$ if $j not=ipm1$ as $x$ and $i$ represent the same thing
$endgroup$
– Henry
Jan 6 at 21:22
add a comment |
$begingroup$
Hints:
If you have $x$ red balls after $n$ picks, what is the probability that your $(n+1)$th pick is red?
or blue?
What happens to the number of red balls in each of these cases?
If you have $x$ red balls after $n$ picks, what are the possible numbers of red balls after $n+1$ picks?
What are the probabilities that you transition to each of these possibilities?
answered Jan 6 at 15:19
HenryHenry
98.8k476163
$endgroup$
Hints:
If you have $x$ red balls after $n$ picks, what is the probability that your $(n+1)$th pick is red?
or blue?
What happens to the number of red balls in each of these cases?
If you have $x$ red balls after $n$ picks, what are the possible numbers of red balls after $n+1$ picks?
What are the probabilities that you transition to each of these possibilities?
answered Jan 6 at 15:19
HenryHenry
98.8k476163
answered Jan 6 at 15:19
HenryHenry
98.8k476163
answered Jan 6 at 15:19
HenryHenry
98.8k476163
answered Jan 6 at 15:19
HenryHenry
98.8k476163
98.8k476163
$begingroup$
okay so far for 1 i got $frac{x}{N}$ and 2 i got $frac{N-x}{N}$ and I have no idea what next.
$endgroup$
– Reety
Jan 6 at 16:19
1
$begingroup$
@Reety You may be overthinking this as you have already done the heavy work. For example with question 3 the number of red balls goes down by $1$ if the $(n+1)$th pick was red, and up by $1$ if the $(n+1)$th pick was blue. So for 4 the number of red balls becomes $x-1$ or $x+1$, and for 5 the respective probabilities of these are those you identified for 1 and 2; these are the transition probabilities as all others are $0$
$endgroup$
– Henry
Jan 6 at 18:17
$begingroup$
$p_{ii}=frac{1}{2}$, $p_{i(i-1)}=frac{x}{N}$, $p_{i(i+1)}=frac{N-x}{N}$ and $p_{ij}=0$ if |i-j|≥2 this is what I got
$endgroup$
– Reety
Jan 6 at 20:42
1
$begingroup$
@Reety I think you should have $p_{ii}=0$ but otherwise that seems correct. Perhaps better to write something like $p_{i,(i-1)}=frac{i}{N}, p_{i,(i+1)}=frac{N-i}{N}, p_{i,j}=0$ if $j not=ipm1$ as $x$ and $i$ represent the same thing
$endgroup$
– Henry
Jan 6 at 21:22
add a comment |
$begingroup$
okay so far for 1 i got $frac{x}{N}$ and 2 i got $frac{N-x}{N}$ and I have no idea what next.
$endgroup$
– Reety
Jan 6 at 16:19
1
$begingroup$
@Reety You may be overthinking this as you have already done the heavy work. For example with question 3 the number of red balls goes down by $1$ if the $(n+1)$th pick was red, and up by $1$ if the $(n+1)$th pick was blue. So for 4 the number of red balls becomes $x-1$ or $x+1$, and for 5 the respective probabilities of these are those you identified for 1 and 2; these are the transition probabilities as all others are $0$
$endgroup$
– Henry
Jan 6 at 18:17
$begingroup$
$p_{ii}=frac{1}{2}$, $p_{i(i-1)}=frac{x}{N}$, $p_{i(i+1)}=frac{N-x}{N}$ and $p_{ij}=0$ if |i-j|≥2 this is what I got
$endgroup$
– Reety
Jan 6 at 20:42
1
$begingroup$
@Reety I think you should have $p_{ii}=0$ but otherwise that seems correct. Perhaps better to write something like $p_{i,(i-1)}=frac{i}{N}, p_{i,(i+1)}=frac{N-i}{N}, p_{i,j}=0$ if $j not=ipm1$ as $x$ and $i$ represent the same thing
$endgroup$
– Henry
Jan 6 at 21:22
$begingroup$
okay so far for 1 i got $frac{x}{N}$ and 2 i got $frac{N-x}{N}$ and I have no idea what next.
$endgroup$
– Reety
Jan 6 at 16:19
$begingroup$
okay so far for 1 i got $frac{x}{N}$ and 2 i got $frac{N-x}{N}$ and I have no idea what next.
$endgroup$
– Reety
Jan 6 at 16:19
1
1
$begingroup$
@Reety You may be overthinking this as you have already done the heavy work. For example with question 3 the number of red balls goes down by $1$ if the $(n+1)$th pick was red, and up by $1$ if the $(n+1)$th pick was blue. So for 4 the number of red balls becomes $x-1$ or $x+1$, and for 5 the respective probabilities of these are those you identified for 1 and 2; these are the transition probabilities as all others are $0$
$endgroup$
– Henry
Jan 6 at 18:17
$begingroup$
@Reety You may be overthinking this as you have already done the heavy work. For example with question 3 the number of red balls goes down by $1$ if the $(n+1)$th pick was red, and up by $1$ if the $(n+1)$th pick was blue. So for 4 the number of red balls becomes $x-1$ or $x+1$, and for 5 the respective probabilities of these are those you identified for 1 and 2; these are the transition probabilities as all others are $0$
$endgroup$
– Henry
Jan 6 at 18:17
$begingroup$
$p_{ii}=frac{1}{2}$, $p_{i(i-1)}=frac{x}{N}$, $p_{i(i+1)}=frac{N-x}{N}$ and $p_{ij}=0$ if |i-j|≥2 this is what I got
$endgroup$
– Reety
Jan 6 at 20:42
$begingroup$
$p_{ii}=frac{1}{2}$, $p_{i(i-1)}=frac{x}{N}$, $p_{i(i+1)}=frac{N-x}{N}$ and $p_{ij}=0$ if |i-j|≥2 this is what I got
$endgroup$
– Reety
Jan 6 at 20:42
1
1
$begingroup$
@Reety I think you should have $p_{ii}=0$ but otherwise that seems correct. Perhaps better to write something like $p_{i,(i-1)}=frac{i}{N}, p_{i,(i+1)}=frac{N-i}{N}, p_{i,j}=0$ if $j not=ipm1$ as $x$ and $i$ represent the same thing
$endgroup$
– Henry
Jan 6 at 21:22
$begingroup$
@Reety I think you should have $p_{ii}=0$ but otherwise that seems correct. Perhaps better to write something like $p_{i,(i-1)}=frac{i}{N}, p_{i,(i+1)}=frac{N-i}{N}, p_{i,j}=0$ if $j not=ipm1$ as $x$ and $i$ represent the same thing
$endgroup$
– Henry
Jan 6 at 21:22
add a comment |
