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So I have the covariance $\mathsf{cov}(x_0,x_1)$ and I know $m_b=−2.5\log_{10}(x_0)+$constant, then how do I calculate $\mathsf{cov}(m_b,x_1)$

I found this but I guess it isn't useful here since it is transformation of both the variables.

Also i tried this $\begin{align}\mathsf{cov}(m_b,x_1) &= \langle\Delta m_b , \Delta x_1 \rangle \\ & = d m_b , d x_1 \\ &= -2.5 /x_0 \, dx_0 , dx_1 \\ &= -2.5 /x_0 \mathsf{cov}(x_0,x_1) \end{align}$

but I am not so sure sure about this method

Martin Sleziak
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SHIN101
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  • Well, the covariance should not be a function of a random variable, so whatever you are doing with the last method is not giving a valid result. – Graham Kemp May 03 '22 at 06:01
  • If all you know is $\mathsf{cov}(x_0,x_1)$ then I do not see how you can find $\mathsf{cov}(m_b,x_1)$. Try a simple example such as $x_0=x_1$ being $1$ or $2$ with equal probability to get $\mathsf{cov}(x_0,x_1)=0.25$ and $\mathsf{cov}(m_b,x_1)\approx -0.188$. Then try $x_0=x_1$ being $11$ or $12$ with equal probability to get $\mathsf{cov}(x_0,x_1)=0.25$ and $\mathsf{cov}(m_b,x_1)\approx -0.024$. – Henry May 03 '22 at 08:29

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