Estimate relationship between two Bernoulli random variables
2
$begingroup$
$X$ and $Y$ are Bernoulli random variables
$X$ and $Y$ are not independent
$x_{t} = P(X_t = 1)$ and $y_{t} = P(X_t = 1)$ for time $t$.- Is it possible to estimate $P(Y = 1 | X = 1)$ from many pairs of $x_{t}$ and $y_{t}$?
I tried:
$$frac{sum_{t}y_{t}x_{t}}{sum_{t}x_{t}}$$
But this formula doesn't make sense because that means $P(X = 1|X = 1)$ is
$$frac{sum_{t}x_{t}x_{t}}{sum_{t}x_{t}}$$
which is not necessarily 1.
probability probability-theory parameter-estimation
asked Dec 12 '18 at 16:49
R zuR zu
33910
$endgroup$
add a comment |
2
$begingroup$
$X$ and $Y$ are Bernoulli random variables
$X$ and $Y$ are not independent
$x_{t} = P(X_t = 1)$ and $y_{t} = P(X_t = 1)$ for time $t$.- Is it possible to estimate $P(Y = 1 | X = 1)$ from many pairs of $x_{t}$ and $y_{t}$?
I tried:
$$frac{sum_{t}y_{t}x_{t}}{sum_{t}x_{t}}$$
But this formula doesn't make sense because that means $P(X = 1|X = 1)$ is
$$frac{sum_{t}x_{t}x_{t}}{sum_{t}x_{t}}$$
which is not necessarily 1.
probability probability-theory parameter-estimation
asked Dec 12 '18 at 16:49
R zuR zu
33910
$endgroup$
$begingroup$
seems related to math.stackexchange.com/questions/610443/…
$endgroup$
– R zu
Dec 12 '18 at 17:00
$begingroup$
Can't do much in this case. Have to assume that the two probability variables are independent at each point of time.
$endgroup$
– R zu
Dec 12 '18 at 18:39
$begingroup$
How do you get $x_t$ and $y_t$? Are they given? Are ${(X_t,Y_t)}$ i.i.d?
$endgroup$
– d.k.o.
Dec 12 '18 at 19:24
$begingroup$
Something here doesn't make too much sense; what is $(X_t,Y_t)$? It seems like you want $(X_t,Y_t)_{t}$ to be independent samples of $(X,Y)$, but their parameters seem to be $(x_t,y_t)$, which further seem to be changing.
$endgroup$
– Marcus M
Dec 12 '18 at 19:25
add a comment |
2
2
2
$begingroup$
$X$ and $Y$ are Bernoulli random variables
$X$ and $Y$ are not independent
$x_{t} = P(X_t = 1)$ and $y_{t} = P(X_t = 1)$ for time $t$.- Is it possible to estimate $P(Y = 1 | X = 1)$ from many pairs of $x_{t}$ and $y_{t}$?
I tried:
$$frac{sum_{t}y_{t}x_{t}}{sum_{t}x_{t}}$$
But this formula doesn't make sense because that means $P(X = 1|X = 1)$ is
$$frac{sum_{t}x_{t}x_{t}}{sum_{t}x_{t}}$$
which is not necessarily 1.
probability probability-theory parameter-estimation
asked Dec 12 '18 at 16:49
R zuR zu
33910
$endgroup$
$X$ and $Y$ are Bernoulli random variables
$X$ and $Y$ are not independent
$x_{t} = P(X_t = 1)$ and $y_{t} = P(X_t = 1)$ for time $t$.- Is it possible to estimate $P(Y = 1 | X = 1)$ from many pairs of $x_{t}$ and $y_{t}$?
I tried:
$$frac{sum_{t}y_{t}x_{t}}{sum_{t}x_{t}}$$
But this formula doesn't make sense because that means $P(X = 1|X = 1)$ is
$$frac{sum_{t}x_{t}x_{t}}{sum_{t}x_{t}}$$
which is not necessarily 1.
probability probability-theory parameter-estimation
probability probability-theory parameter-estimation
asked Dec 12 '18 at 16:49
R zuR zu
33910
asked Dec 12 '18 at 16:49
R zuR zu
33910
edited Dec 12 '18 at 17:23
R zu
asked Dec 12 '18 at 16:49
R zuR zu
33910
asked Dec 12 '18 at 16:49
R zuR zu
33910
asked Dec 12 '18 at 16:49
R zuR zu
33910
33910
$begingroup$
seems related to math.stackexchange.com/questions/610443/…
$endgroup$
– R zu
Dec 12 '18 at 17:00
$begingroup$
Can't do much in this case. Have to assume that the two probability variables are independent at each point of time.
$endgroup$
– R zu
Dec 12 '18 at 18:39
$begingroup$
How do you get $x_t$ and $y_t$? Are they given? Are ${(X_t,Y_t)}$ i.i.d?
$endgroup$
– d.k.o.
Dec 12 '18 at 19:24
$begingroup$
Something here doesn't make too much sense; what is $(X_t,Y_t)$? It seems like you want $(X_t,Y_t)_{t}$ to be independent samples of $(X,Y)$, but their parameters seem to be $(x_t,y_t)$, which further seem to be changing.
$endgroup$
– Marcus M
Dec 12 '18 at 19:25
add a comment |
$begingroup$
seems related to math.stackexchange.com/questions/610443/…
$endgroup$
– R zu
Dec 12 '18 at 17:00
$begingroup$
Can't do much in this case. Have to assume that the two probability variables are independent at each point of time.
$endgroup$
– R zu
Dec 12 '18 at 18:39
$begingroup$
How do you get $x_t$ and $y_t$? Are they given? Are ${(X_t,Y_t)}$ i.i.d?
$endgroup$
– d.k.o.
Dec 12 '18 at 19:24
$begingroup$
Something here doesn't make too much sense; what is $(X_t,Y_t)$? It seems like you want $(X_t,Y_t)_{t}$ to be independent samples of $(X,Y)$, but their parameters seem to be $(x_t,y_t)$, which further seem to be changing.
$endgroup$
– Marcus M
Dec 12 '18 at 19:25
$begingroup$
seems related to math.stackexchange.com/questions/610443/…
$endgroup$
– R zu
Dec 12 '18 at 17:00
$begingroup$
seems related to math.stackexchange.com/questions/610443/…
$endgroup$
– R zu
Dec 12 '18 at 17:00
$begingroup$
Can't do much in this case. Have to assume that the two probability variables are independent at each point of time.
$endgroup$
– R zu
Dec 12 '18 at 18:39
$begingroup$
Can't do much in this case. Have to assume that the two probability variables are independent at each point of time.
$endgroup$
– R zu
Dec 12 '18 at 18:39
$begingroup$
How do you get $x_t$ and $y_t$? Are they given? Are ${(X_t,Y_t)}$ i.i.d?
$endgroup$
– d.k.o.
Dec 12 '18 at 19:24
$begingroup$
How do you get $x_t$ and $y_t$? Are they given? Are ${(X_t,Y_t)}$ i.i.d?
$endgroup$
– d.k.o.
Dec 12 '18 at 19:24
$begingroup$
Something here doesn't make too much sense; what is $(X_t,Y_t)$? It seems like you want $(X_t,Y_t)_{t}$ to be independent samples of $(X,Y)$, but their parameters seem to be $(x_t,y_t)$, which further seem to be changing.
$endgroup$
– Marcus M
Dec 12 '18 at 19:25
$begingroup$
Something here doesn't make too much sense; what is $(X_t,Y_t)$? It seems like you want $(X_t,Y_t)_{t}$ to be independent samples of $(X,Y)$, but their parameters seem to be $(x_t,y_t)$, which further seem to be changing.
$endgroup$
– Marcus M
Dec 12 '18 at 19:25
add a comment |
1 Answer
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$begingroup$
Throw away the data where $x_t neq 1$. Then compute the proportion of the remaining data where $y_t = 1$.
answered Dec 13 '18 at 3:39
hunterhunter
15.5k32640
$endgroup$
add a comment |
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$begingroup$
Throw away the data where $x_t neq 1$. Then compute the proportion of the remaining data where $y_t = 1$.
answered Dec 13 '18 at 3:39
hunterhunter
15.5k32640
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1
$begingroup$
Throw away the data where $x_t neq 1$. Then compute the proportion of the remaining data where $y_t = 1$.
answered Dec 13 '18 at 3:39
hunterhunter
15.5k32640
$endgroup$
add a comment |
1
1
1
$begingroup$
Throw away the data where $x_t neq 1$. Then compute the proportion of the remaining data where $y_t = 1$.
answered Dec 13 '18 at 3:39
hunterhunter
15.5k32640
$endgroup$
Throw away the data where $x_t neq 1$. Then compute the proportion of the remaining data where $y_t = 1$.
answered Dec 13 '18 at 3:39
hunterhunter
15.5k32640
answered Dec 13 '18 at 3:39
hunterhunter
15.5k32640
answered Dec 13 '18 at 3:39
hunterhunter
15.5k32640
answered Dec 13 '18 at 3:39
hunterhunter
15.5k32640
15.5k32640
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add a comment |
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$begingroup$
seems related to math.stackexchange.com/questions/610443/…
$endgroup$
– R zu
Dec 12 '18 at 17:00
$begingroup$
Can't do much in this case. Have to assume that the two probability variables are independent at each point of time.
$endgroup$
– R zu
Dec 12 '18 at 18:39
$begingroup$
How do you get $x_t$ and $y_t$? Are they given? Are ${(X_t,Y_t)}$ i.i.d?
$endgroup$
– d.k.o.
Dec 12 '18 at 19:24
$begingroup$
Something here doesn't make too much sense; what is $(X_t,Y_t)$? It seems like you want $(X_t,Y_t)_{t}$ to be independent samples of $(X,Y)$, but their parameters seem to be $(x_t,y_t)$, which further seem to be changing.
$endgroup$
– Marcus M
Dec 12 '18 at 19:25