expectation of number of exactly 3 heads flip in a row
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lets say I have 400 coin tosses and a probability P for head.
${ x_n }$ the tosses,then X is a random variable which X = {number of times which there where exactly 3 heads}
how can I calculate E[x] and Var[x]?
I tried to get to the solution through recursion by choose where the first 3 exactly heads will be but with no luck.
k=400 ,
P(X=n) = F(n,k) = F(n-1,k-4)+F(n-1,k-5)+F(n-1,k-6)+F(n-1,k-7)
probability sequences-and-series probability-theory stochastic-processes
asked Jan 3 at 18:58
MSmMSm
35719
$endgroup$
add a comment |
$begingroup$
lets say I have 400 coin tosses and a probability P for head.
${ x_n }$ the tosses,then X is a random variable which X = {number of times which there where exactly 3 heads}
how can I calculate E[x] and Var[x]?
I tried to get to the solution through recursion by choose where the first 3 exactly heads will be but with no luck.
k=400 ,
P(X=n) = F(n,k) = F(n-1,k-4)+F(n-1,k-5)+F(n-1,k-6)+F(n-1,k-7)
probability sequences-and-series probability-theory stochastic-processes
asked Jan 3 at 18:58
MSmMSm
35719
$endgroup$
2
$begingroup$
Hint (for expected value at least): use Linearity of Expectation, and for $iin {1,398}$ let $X_i$ be the indicator variable which tells you if a string of exactly three Heads starts with the $i^{th}$ toss.
$endgroup$
– lulu
Jan 3 at 19:00
$begingroup$
but $X neq sum_i X_i$
$endgroup$
– MSm
Jan 4 at 0:30
$begingroup$
Not following. Why do you think they aren't equal? The number of exact triples $HHH$ equals the number of places where exact triples start. Note: Variance can be done this way but it is more work. For each pair $i,j$ you have to analyze $E[X_iX_j]$.
$endgroup$
– lulu
Jan 4 at 11:15
$begingroup$
Ok Now I get it, thanks!
$endgroup$
– MSm
Jan 4 at 11:55
add a comment |
$begingroup$
lets say I have 400 coin tosses and a probability P for head.
${ x_n }$ the tosses,then X is a random variable which X = {number of times which there where exactly 3 heads}
how can I calculate E[x] and Var[x]?
I tried to get to the solution through recursion by choose where the first 3 exactly heads will be but with no luck.
k=400 ,
P(X=n) = F(n,k) = F(n-1,k-4)+F(n-1,k-5)+F(n-1,k-6)+F(n-1,k-7)
probability sequences-and-series probability-theory stochastic-processes
asked Jan 3 at 18:58
MSmMSm
35719
$endgroup$
lets say I have 400 coin tosses and a probability P for head.
${ x_n }$ the tosses,then X is a random variable which X = {number of times which there where exactly 3 heads}
how can I calculate E[x] and Var[x]?
I tried to get to the solution through recursion by choose where the first 3 exactly heads will be but with no luck.
k=400 ,
P(X=n) = F(n,k) = F(n-1,k-4)+F(n-1,k-5)+F(n-1,k-6)+F(n-1,k-7)
probability sequences-and-series probability-theory stochastic-processes
probability sequences-and-series probability-theory stochastic-processes
asked Jan 3 at 18:58
MSmMSm
35719
asked Jan 3 at 18:58
MSmMSm
35719
asked Jan 3 at 18:58
MSmMSm
35719
asked Jan 3 at 18:58
MSmMSm
35719
asked Jan 3 at 18:58
MSmMSm
35719
35719
2
$begingroup$
Hint (for expected value at least): use Linearity of Expectation, and for $iin {1,398}$ let $X_i$ be the indicator variable which tells you if a string of exactly three Heads starts with the $i^{th}$ toss.
$endgroup$
– lulu
Jan 3 at 19:00
$begingroup$
but $X neq sum_i X_i$
$endgroup$
– MSm
Jan 4 at 0:30
$begingroup$
Not following. Why do you think they aren't equal? The number of exact triples $HHH$ equals the number of places where exact triples start. Note: Variance can be done this way but it is more work. For each pair $i,j$ you have to analyze $E[X_iX_j]$.
$endgroup$
– lulu
Jan 4 at 11:15
$begingroup$
Ok Now I get it, thanks!
$endgroup$
– MSm
Jan 4 at 11:55
add a comment |
2
$begingroup$
Hint (for expected value at least): use Linearity of Expectation, and for $iin {1,398}$ let $X_i$ be the indicator variable which tells you if a string of exactly three Heads starts with the $i^{th}$ toss.
$endgroup$
– lulu
Jan 3 at 19:00
$begingroup$
but $X neq sum_i X_i$
$endgroup$
– MSm
Jan 4 at 0:30
$begingroup$
Not following. Why do you think they aren't equal? The number of exact triples $HHH$ equals the number of places where exact triples start. Note: Variance can be done this way but it is more work. For each pair $i,j$ you have to analyze $E[X_iX_j]$.
$endgroup$
– lulu
Jan 4 at 11:15
$begingroup$
Ok Now I get it, thanks!
$endgroup$
– MSm
Jan 4 at 11:55
2
2
$begingroup$
Hint (for expected value at least): use Linearity of Expectation, and for $iin {1,398}$ let $X_i$ be the indicator variable which tells you if a string of exactly three Heads starts with the $i^{th}$ toss.
$endgroup$
– lulu
Jan 3 at 19:00
$begingroup$
Hint (for expected value at least): use Linearity of Expectation, and for $iin {1,398}$ let $X_i$ be the indicator variable which tells you if a string of exactly three Heads starts with the $i^{th}$ toss.
$endgroup$
– lulu
Jan 3 at 19:00
$begingroup$
but $X neq sum_i X_i$
$endgroup$
– MSm
Jan 4 at 0:30
$begingroup$
but $X neq sum_i X_i$
$endgroup$
– MSm
Jan 4 at 0:30
$begingroup$
Not following. Why do you think they aren't equal? The number of exact triples $HHH$ equals the number of places where exact triples start. Note: Variance can be done this way but it is more work. For each pair $i,j$ you have to analyze $E[X_iX_j]$.
$endgroup$
– lulu
Jan 4 at 11:15
$begingroup$
Not following. Why do you think they aren't equal? The number of exact triples $HHH$ equals the number of places where exact triples start. Note: Variance can be done this way but it is more work. For each pair $i,j$ you have to analyze $E[X_iX_j]$.
$endgroup$
– lulu
Jan 4 at 11:15
$begingroup$
Ok Now I get it, thanks!
$endgroup$
– MSm
Jan 4 at 11:55
$begingroup$
Ok Now I get it, thanks!
$endgroup$
– MSm
Jan 4 at 11:55
add a comment |
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$begingroup$
Hint (for expected value at least): use Linearity of Expectation, and for $iin {1,398}$ let $X_i$ be the indicator variable which tells you if a string of exactly three Heads starts with the $i^{th}$ toss.
$endgroup$
– lulu
Jan 3 at 19:00
$begingroup$
but $X neq sum_i X_i$
$endgroup$
– MSm
Jan 4 at 0:30
$begingroup$
Not following. Why do you think they aren't equal? The number of exact triples $HHH$ equals the number of places where exact triples start. Note: Variance can be done this way but it is more work. For each pair $i,j$ you have to analyze $E[X_iX_j]$.
$endgroup$
– lulu
Jan 4 at 11:15
$begingroup$
Ok Now I get it, thanks!
$endgroup$
– MSm
Jan 4 at 11:55