Can $P(X<Y)$ be calculated like this in the case of correlated measurements?
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I have two methods, $mathcalX$ and $mathcalY$, and want to determine which one produces lower responses. I figured to estimate $P(X<Y)$ where $X$ is the random variable representing the responses of method $mathcalX$, whereas $Y$ denotes the responses of method $mathcalY$. Now, the methods have been tested on the same group of $l$ subjects for $n$ times. This means that for each subject $s_i$ I have the responses $x_i,1,x_i,2,ldots,x_i,n$ for method $mathcalX$ and $y_i,1,y_i,2,ldots,y_i,n$ for $mathcalY$.
I know that I can estimate $P(X<Y)$ for a particular subject $s_i$ as follows:
$$p_i = fracU_in^2$$
where $U_i$ is the Mann-Whitney U statistic for $x_i,1,x_i,2,ldots,x_i,n$ and $y_i,1,y_i,2,ldots,y_i,n$.
My question is:
Is it correct to estimate $P(X < Y)$ as the average of $p_i$, i.e. as follows?$$frac1lsum_i=1^lp_i=frac1lsum_i=1^lfracU_in^2=frac1n^2lsum_i=1^lU_i$$
probability statistics
asked Aug 24 at 2:44
Milos
1275
add a comment |Â
up vote
1
down vote
favorite
I have two methods, $mathcalX$ and $mathcalY$, and want to determine which one produces lower responses. I figured to estimate $P(X<Y)$ where $X$ is the random variable representing the responses of method $mathcalX$, whereas $Y$ denotes the responses of method $mathcalY$. Now, the methods have been tested on the same group of $l$ subjects for $n$ times. This means that for each subject $s_i$ I have the responses $x_i,1,x_i,2,ldots,x_i,n$ for method $mathcalX$ and $y_i,1,y_i,2,ldots,y_i,n$ for $mathcalY$.
I know that I can estimate $P(X<Y)$ for a particular subject $s_i$ as follows:
$$p_i = fracU_in^2$$
where $U_i$ is the Mann-Whitney U statistic for $x_i,1,x_i,2,ldots,x_i,n$ and $y_i,1,y_i,2,ldots,y_i,n$.
My question is:
Is it correct to estimate $P(X < Y)$ as the average of $p_i$, i.e. as follows?$$frac1lsum_i=1^lp_i=frac1lsum_i=1^lfracU_in^2=frac1n^2lsum_i=1^lU_i$$
probability statistics
asked Aug 24 at 2:44
Milos
1275
Also posted to stats
– Glen_b
Aug 25 at 3:30
That cross-post to stats seems to have been deleted in the meantime.
– joriki
Aug 31 at 22:00
Perhaps you could add a bit about what makes you think that this might not be correct?
– joriki
Aug 31 at 22:01
Well, I'm not sure if the mean of individual probabilities $P(X<Y)$ is the probability that $X<Y$ for a random individual.
– Milos
Sep 1 at 6:30
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have two methods, $mathcalX$ and $mathcalY$, and want to determine which one produces lower responses. I figured to estimate $P(X<Y)$ where $X$ is the random variable representing the responses of method $mathcalX$, whereas $Y$ denotes the responses of method $mathcalY$. Now, the methods have been tested on the same group of $l$ subjects for $n$ times. This means that for each subject $s_i$ I have the responses $x_i,1,x_i,2,ldots,x_i,n$ for method $mathcalX$ and $y_i,1,y_i,2,ldots,y_i,n$ for $mathcalY$.
I know that I can estimate $P(X<Y)$ for a particular subject $s_i$ as follows:
$$p_i = fracU_in^2$$
where $U_i$ is the Mann-Whitney U statistic for $x_i,1,x_i,2,ldots,x_i,n$ and $y_i,1,y_i,2,ldots,y_i,n$.
My question is:
Is it correct to estimate $P(X < Y)$ as the average of $p_i$, i.e. as follows?$$frac1lsum_i=1^lp_i=frac1lsum_i=1^lfracU_in^2=frac1n^2lsum_i=1^lU_i$$
probability statistics
asked Aug 24 at 2:44
Milos
1275
I have two methods, $mathcalX$ and $mathcalY$, and want to determine which one produces lower responses. I figured to estimate $P(X<Y)$ where $X$ is the random variable representing the responses of method $mathcalX$, whereas $Y$ denotes the responses of method $mathcalY$. Now, the methods have been tested on the same group of $l$ subjects for $n$ times. This means that for each subject $s_i$ I have the responses $x_i,1,x_i,2,ldots,x_i,n$ for method $mathcalX$ and $y_i,1,y_i,2,ldots,y_i,n$ for $mathcalY$.
I know that I can estimate $P(X<Y)$ for a particular subject $s_i$ as follows:
$$p_i = fracU_in^2$$
where $U_i$ is the Mann-Whitney U statistic for $x_i,1,x_i,2,ldots,x_i,n$ and $y_i,1,y_i,2,ldots,y_i,n$.
My question is:
Is it correct to estimate $P(X < Y)$ as the average of $p_i$, i.e. as follows?$$frac1lsum_i=1^lp_i=frac1lsum_i=1^lfracU_in^2=frac1n^2lsum_i=1^lU_i$$
probability statistics
asked Aug 24 at 2:44
Milos
1275
edited Aug 31 at 19:51
asked Aug 24 at 2:44
Milos
1275
asked Aug 24 at 2:44
Milos
1275
asked Aug 24 at 2:44
Milos
1275
1275
Also posted to stats
– Glen_b
Aug 25 at 3:30
That cross-post to stats seems to have been deleted in the meantime.
– joriki
Aug 31 at 22:00
Perhaps you could add a bit about what makes you think that this might not be correct?
– joriki
Aug 31 at 22:01
Well, I'm not sure if the mean of individual probabilities $P(X<Y)$ is the probability that $X<Y$ for a random individual.
– Milos
Sep 1 at 6:30
add a comment |Â
Also posted to stats
– Glen_b
Aug 25 at 3:30
That cross-post to stats seems to have been deleted in the meantime.
– joriki
Aug 31 at 22:00
Perhaps you could add a bit about what makes you think that this might not be correct?
– joriki
Aug 31 at 22:01
Well, I'm not sure if the mean of individual probabilities $P(X<Y)$ is the probability that $X<Y$ for a random individual.
– Milos
Sep 1 at 6:30
Also posted to stats
– Glen_b
Aug 25 at 3:30
Also posted to stats
– Glen_b
Aug 25 at 3:30
That cross-post to stats seems to have been deleted in the meantime.
– joriki
Aug 31 at 22:00
That cross-post to stats seems to have been deleted in the meantime.
– joriki
Aug 31 at 22:00
Perhaps you could add a bit about what makes you think that this might not be correct?
– joriki
Aug 31 at 22:01
Perhaps you could add a bit about what makes you think that this might not be correct?
– joriki
Aug 31 at 22:01
Well, I'm not sure if the mean of individual probabilities $P(X<Y)$ is the probability that $X<Y$ for a random individual.
– Milos
Sep 1 at 6:30
Well, I'm not sure if the mean of individual probabilities $P(X<Y)$ is the probability that $X<Y$ for a random individual.
– Milos
Sep 1 at 6:30
add a comment |Â
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Also posted to stats
– Glen_b
Aug 25 at 3:30
That cross-post to stats seems to have been deleted in the meantime.
– joriki
Aug 31 at 22:00
Perhaps you could add a bit about what makes you think that this might not be correct?
– joriki
Aug 31 at 22:01
Well, I'm not sure if the mean of individual probabilities $P(X<Y)$ is the probability that $X<Y$ for a random individual.
– Milos
Sep 1 at 6:30