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?$$frac1lsum_i=1^lp_i=frac1lsum_i=1^lfracU_in^2=frac1n^2lsum_i=1^lU_i$$







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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














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?$$frac1lsum_i=1^lp_i=frac1lsum_i=1^lfracU_in^2=frac1n^2lsum_i=1^lU_i$$







share|cite|improve this question






















  • 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












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?$$frac1lsum_i=1^lp_i=frac1lsum_i=1^lfracU_in^2=frac1n^2lsum_i=1^lU_i$$







share|cite|improve this question














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?$$frac1lsum_i=1^lp_i=frac1lsum_i=1^lfracU_in^2=frac1n^2lsum_i=1^lU_i$$









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 31 at 19:51

























asked Aug 24 at 2:44









Milos

1275




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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
















  • 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















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