Mixing
Find the probability mass function of the waiting time to get $TTHH$?
$begingroup$
Suppose we are flipping a fair coin. Let $X$ be the waiting to the first occurrence of $TTHH$ (Tails, Tails, Heads, Heads). So if we got $HTTHH$, then $X=5$. Find the pmf of $X$.
Now I've tried conditioning on the first flip ($X_1$) but I get nowhere:
$$begin{align}
P(X=n) &= frac{1}{2}P(X=n|X_1=T)+frac{1}{2}P(X=n|X_1=H) \
&=frac{1}{2}P(X=n|X_1=T)+frac{1}{2}P(X=n-1) \
end{align}$$
I'm not sure where to go from here or even if this is the right approach. I've even tried to mess around with Markov Chains but can't seem to get anywhere.
probability conditional-probability
asked Feb 2 at 19:22
NineteenNinetyNineNineteenNinetyNine
769
$endgroup$
add a comment |
$begingroup$
Suppose we are flipping a fair coin. Let $X$ be the waiting to the first occurrence of $TTHH$ (Tails, Tails, Heads, Heads). So if we got $HTTHH$, then $X=5$. Find the pmf of $X$.
Now I've tried conditioning on the first flip ($X_1$) but I get nowhere:
$$begin{align}
P(X=n) &= frac{1}{2}P(X=n|X_1=T)+frac{1}{2}P(X=n|X_1=H) \
&=frac{1}{2}P(X=n|X_1=T)+frac{1}{2}P(X=n-1) \
end{align}$$
I'm not sure where to go from here or even if this is the right approach. I've even tried to mess around with Markov Chains but can't seem to get anywhere.
probability conditional-probability
asked Feb 2 at 19:22
NineteenNinetyNineNineteenNinetyNine
769
$endgroup$
1
$begingroup$
I doubt there is a simple closed form for the probability mass function, but the probability generating function may be $G(z)=dfrac{z^4}{z^4-16z+16}$
$endgroup$
– Henry
Feb 2 at 19:57
$begingroup$
Yes that agrees with my result. How did you find the PGF?
$endgroup$
– Peter Foreman
Feb 2 at 20:28
add a comment |
$begingroup$
Suppose we are flipping a fair coin. Let $X$ be the waiting to the first occurrence of $TTHH$ (Tails, Tails, Heads, Heads). So if we got $HTTHH$, then $X=5$. Find the pmf of $X$.
Now I've tried conditioning on the first flip ($X_1$) but I get nowhere:
$$begin{align}
P(X=n) &= frac{1}{2}P(X=n|X_1=T)+frac{1}{2}P(X=n|X_1=H) \
&=frac{1}{2}P(X=n|X_1=T)+frac{1}{2}P(X=n-1) \
end{align}$$
I'm not sure where to go from here or even if this is the right approach. I've even tried to mess around with Markov Chains but can't seem to get anywhere.
probability conditional-probability
asked Feb 2 at 19:22
NineteenNinetyNineNineteenNinetyNine
769
$endgroup$
Suppose we are flipping a fair coin. Let $X$ be the waiting to the first occurrence of $TTHH$ (Tails, Tails, Heads, Heads). So if we got $HTTHH$, then $X=5$. Find the pmf of $X$.
Now I've tried conditioning on the first flip ($X_1$) but I get nowhere:
$$begin{align}
P(X=n) &= frac{1}{2}P(X=n|X_1=T)+frac{1}{2}P(X=n|X_1=H) \
&=frac{1}{2}P(X=n|X_1=T)+frac{1}{2}P(X=n-1) \
end{align}$$
I'm not sure where to go from here or even if this is the right approach. I've even tried to mess around with Markov Chains but can't seem to get anywhere.
probability conditional-probability
probability conditional-probability
asked Feb 2 at 19:22
NineteenNinetyNineNineteenNinetyNine
769
asked Feb 2 at 19:22
NineteenNinetyNineNineteenNinetyNine
769
asked Feb 2 at 19:22
NineteenNinetyNineNineteenNinetyNine
769
asked Feb 2 at 19:22
NineteenNinetyNineNineteenNinetyNine
769
asked Feb 2 at 19:22
NineteenNinetyNineNineteenNinetyNine
769
769
1
$begingroup$
I doubt there is a simple closed form for the probability mass function, but the probability generating function may be $G(z)=dfrac{z^4}{z^4-16z+16}$
$endgroup$
– Henry
Feb 2 at 19:57
$begingroup$
Yes that agrees with my result. How did you find the PGF?
$endgroup$
– Peter Foreman
Feb 2 at 20:28
add a comment |
1
$begingroup$
I doubt there is a simple closed form for the probability mass function, but the probability generating function may be $G(z)=dfrac{z^4}{z^4-16z+16}$
$endgroup$
– Henry
Feb 2 at 19:57
$begingroup$
Yes that agrees with my result. How did you find the PGF?
$endgroup$
– Peter Foreman
Feb 2 at 20:28
1
1
$begingroup$
I doubt there is a simple closed form for the probability mass function, but the probability generating function may be $G(z)=dfrac{z^4}{z^4-16z+16}$
$endgroup$
– Henry
Feb 2 at 19:57
$begingroup$
I doubt there is a simple closed form for the probability mass function, but the probability generating function may be $G(z)=dfrac{z^4}{z^4-16z+16}$
$endgroup$
– Henry
Feb 2 at 19:57
$begingroup$
Yes that agrees with my result. How did you find the PGF?
$endgroup$
– Peter Foreman
Feb 2 at 20:28
$begingroup$
Yes that agrees with my result. How did you find the PGF?
$endgroup$
– Peter Foreman
Feb 2 at 20:28
add a comment |
1 Answer
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oldest
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$begingroup$
One can easily see that $P(X le 3)=0$ and $P(X=4) = frac{1}{2^4}=frac{1}{16}$. To get $P(X=5)$ we need to get TTHH in the last 4 throws and to not get TTHH on the first roll, so $P(X=5)=frac{1}{16}(1-P(X=1))=frac{1}{16}$. Similarly $P(X=6)=frac{1}{16}(1-P(X le 2))=frac{1}{16}$ and $P(X=7)=frac{1}{16}(1-P(X le 3))=frac{1}{16}$. After 8 throws we have $P(X=8)=frac{1}{16}(1-P(X le 4))=frac{1}{16}(1-frac{1}{16})=frac{15}{256}$.
In general we have:
$$P(X=x)=frac{1}{16}(1-P(X le (x-1)))=frac{1}{16}-frac{1}{16}sum_{k=0}^{x-1}P(X=k)$$
I don't know how to find a closed form for this answer.
answered Feb 2 at 20:22
Peter ForemanPeter Foreman
7,2711319
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
One can easily see that $P(X le 3)=0$ and $P(X=4) = frac{1}{2^4}=frac{1}{16}$. To get $P(X=5)$ we need to get TTHH in the last 4 throws and to not get TTHH on the first roll, so $P(X=5)=frac{1}{16}(1-P(X=1))=frac{1}{16}$. Similarly $P(X=6)=frac{1}{16}(1-P(X le 2))=frac{1}{16}$ and $P(X=7)=frac{1}{16}(1-P(X le 3))=frac{1}{16}$. After 8 throws we have $P(X=8)=frac{1}{16}(1-P(X le 4))=frac{1}{16}(1-frac{1}{16})=frac{15}{256}$.
In general we have:
$$P(X=x)=frac{1}{16}(1-P(X le (x-1)))=frac{1}{16}-frac{1}{16}sum_{k=0}^{x-1}P(X=k)$$
I don't know how to find a closed form for this answer.
answered Feb 2 at 20:22
Peter ForemanPeter Foreman
7,2711319
$endgroup$
add a comment |
$begingroup$
One can easily see that $P(X le 3)=0$ and $P(X=4) = frac{1}{2^4}=frac{1}{16}$. To get $P(X=5)$ we need to get TTHH in the last 4 throws and to not get TTHH on the first roll, so $P(X=5)=frac{1}{16}(1-P(X=1))=frac{1}{16}$. Similarly $P(X=6)=frac{1}{16}(1-P(X le 2))=frac{1}{16}$ and $P(X=7)=frac{1}{16}(1-P(X le 3))=frac{1}{16}$. After 8 throws we have $P(X=8)=frac{1}{16}(1-P(X le 4))=frac{1}{16}(1-frac{1}{16})=frac{15}{256}$.
In general we have:
$$P(X=x)=frac{1}{16}(1-P(X le (x-1)))=frac{1}{16}-frac{1}{16}sum_{k=0}^{x-1}P(X=k)$$
I don't know how to find a closed form for this answer.
answered Feb 2 at 20:22
Peter ForemanPeter Foreman
7,2711319
$endgroup$
add a comment |
$begingroup$
One can easily see that $P(X le 3)=0$ and $P(X=4) = frac{1}{2^4}=frac{1}{16}$. To get $P(X=5)$ we need to get TTHH in the last 4 throws and to not get TTHH on the first roll, so $P(X=5)=frac{1}{16}(1-P(X=1))=frac{1}{16}$. Similarly $P(X=6)=frac{1}{16}(1-P(X le 2))=frac{1}{16}$ and $P(X=7)=frac{1}{16}(1-P(X le 3))=frac{1}{16}$. After 8 throws we have $P(X=8)=frac{1}{16}(1-P(X le 4))=frac{1}{16}(1-frac{1}{16})=frac{15}{256}$.
In general we have:
$$P(X=x)=frac{1}{16}(1-P(X le (x-1)))=frac{1}{16}-frac{1}{16}sum_{k=0}^{x-1}P(X=k)$$
I don't know how to find a closed form for this answer.
answered Feb 2 at 20:22
Peter ForemanPeter Foreman
7,2711319
$endgroup$
One can easily see that $P(X le 3)=0$ and $P(X=4) = frac{1}{2^4}=frac{1}{16}$. To get $P(X=5)$ we need to get TTHH in the last 4 throws and to not get TTHH on the first roll, so $P(X=5)=frac{1}{16}(1-P(X=1))=frac{1}{16}$. Similarly $P(X=6)=frac{1}{16}(1-P(X le 2))=frac{1}{16}$ and $P(X=7)=frac{1}{16}(1-P(X le 3))=frac{1}{16}$. After 8 throws we have $P(X=8)=frac{1}{16}(1-P(X le 4))=frac{1}{16}(1-frac{1}{16})=frac{15}{256}$.
In general we have:
$$P(X=x)=frac{1}{16}(1-P(X le (x-1)))=frac{1}{16}-frac{1}{16}sum_{k=0}^{x-1}P(X=k)$$
I don't know how to find a closed form for this answer.
answered Feb 2 at 20:22
Peter ForemanPeter Foreman
7,2711319
answered Feb 2 at 20:22
Peter ForemanPeter Foreman
7,2711319
answered Feb 2 at 20:22
Peter ForemanPeter Foreman
7,2711319
answered Feb 2 at 20:22
Peter ForemanPeter Foreman
7,2711319
7,2711319
add a comment |
add a comment |
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$begingroup$
I doubt there is a simple closed form for the probability mass function, but the probability generating function may be $G(z)=dfrac{z^4}{z^4-16z+16}$
$endgroup$
– Henry
Feb 2 at 19:57
$begingroup$
Yes that agrees with my result. How did you find the PGF?
$endgroup$
– Peter Foreman
Feb 2 at 20:28