Correlation boundLet x and y be two random variables such that:Corr(x,y) = b, where Corr(x,y) represents correlation between x and y, b is a scalar number in range of [-1, 1]. Let y' be an estimation of y. An example could be y'=y+(rand(0,1)-0.5)*.1, rand(0,1) gives random number between 0, 1. I am adding some noise to the data.My questions are:Is there a way where I can bound the correlation between x, y' i.e. Corr(x,y')?I mentioned y' in light of random perturbation, I would like to know what if I don't have that information, where I only know that y' is a estimation of y. Are there any literature that cover it?

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Correlation boundLet x and y be two random variables such that:Corr(x,y) = b, where Corr(x,y) represents correlation between x and y, b is a scalar number in range of [-1, 1]. Let y&#039; be an estimation of y. An example could be y&#039;=y+(rand(0,1)-0.5)*.1, rand(0,1) gives random number between 0, 1. I am adding some noise to the data.My questions are:Is there a way where I can bound the correlation between x, y&#039; i.e. Corr(x,y&#039;)?I mentioned y&#039; in light of random perturbation, I would like to know what if I don&#039;t have that information, where I only know that y&#039; is a estimation of y. Are there any literature that cover it?

Julius Haley · Accepted Answer

Let   e  =      y    ′    −  y. Assuming that e is independent from x and y with       μ    e    =  E  (  e  )  =  0, then       μ                  y        ′              =  E  (      y    ′    )  =  E  (  y  )  =      μ    y   and:                              C          o          r          r                (        x        ,                  y          ′                )                            =                              E            (            (            x            −                          μ              x                        )            (                          y              ′                        −                          μ              y                        )            )                                              σ              x                                      σ                                                y                  ′                                                                                                      =                              E            (            (            x            −                          μ              x                        )            (            y            −                          μ              y                        )            )            +            E            (            (            x            −                          μ              x                        )            e            )                                              σ              x                                      σ              y              ′                                                                        =                  C          o          r          r                (        x        ,        y        )                              σ            y                                σ                                          y                ′                                                          E  (  (  x  −      μ    x    )  e  )  =  E  (  x  −      μ    x    )  E  (  e  )  =  0 since x and e are independent.Now,       σ                  y        ′              =            σ      y      2        +          σ      e      2      , again by independence, so:      C    o    r    r    (  x  ,      y    ′    )  =  C  o  r  r  (  x  ,  y  )      1          1      +                        (                                    σ              e                                      σ              y                                )                2            So definitely       |        C    o    r    r    (  x  ,      y    ′    )      |    &amp;lt;      |        C    o    r    r    (  x  ,  y  )      |  I believe the specific e you have given, we have       σ    e    =      0.1          6      There is no meaning to &quot;estimation&quot; technically. You can always say that   y  −      y    ′   is another random variable. If you don&#039;t know that y′−y is independent of x, you don&#039;t know what   E  (  (  x  −      μ    x    )  (  e  −      μ    e    )  ) is. If you don&#039;t know that   e  =      y    ′    −  y and y are independent, you don&#039;t know       σ                  y        ′             in terms of       σ          e       and       σ          y      . In particular, you don&#039;t know       σ                  y        ′              &amp;gt;      σ          y      .A simple example is that if       y    ′    =  x then   E  (  x  ,      y    ′    )  =  1. So if x,y are close enough that x can be said to be an &quot;estimate for y&quot; then   E  (  x  ,  x  )  =  1  &amp;gt;  E  (  x  ,  y  )).

mxty42ued · Accepted Answer

Let X,Y be random variables with a given correlation b. Let Z be any random variable independent of σ(X,Y), and suppose   Z  ≠  0 has a strictly positive, finite variance, and       E    [  Z  ]  =  0. Here Z is &#039;noise&#039; that will contribute to       Y    ′    =  Y  +  ZNotice that   C  o  v  (  X  ,  Z  )  =  0 and   C  o  v  (  X  ,      Y    ′    )  =  C  o  v  (  X  ,  Y  )  +  C  o  v  (  X  ,  Z  )  =  C  o  v  (  X  ,  Y  )since Z is independent of X. Moreover   V  a  r  (      Y    ′    )  =  V  a  r  (  Y  )  +  V  a  r  (  Z  )  &amp;gt;  V  a  r  (  Y  ) by independence. We conclude that      |    C  o  r  r  (  X  ,      Y    ′    )      |    =      |                  C        o        v        (        X        ,                  Y          ′                )                    V        a        r        (        X        )        V        a        r        (                  Y          ′                )              |    &amp;lt;      |                  C        o        v        (        X        ,        Y        )                    V        a        r        (        X        )        V        a        r        (        Y        )              |    =      |    C  o  r  r  (  X  ,  Y  )      |  We conclude that the addition of any zero-expectation, independent noise of finite variance will diminish the correlation.

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