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If given:

$$cov(x, y) = 0, cov(x, z) = 0$$

then can we conclude that:

$$cov(y, z) = 0$$

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There is a kind of stupid counter-example. If $y = z$ then your conditions degenerate to one condition, and are asking if $\text{cov}(x,y)=0 \implies \text{cov}(y,y)=0$, but this certainly need not be true!

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No. If $X, Y$ are independent and $Y, Z$ are independent, then $X, Z$ does not have to be independent. For example, if $X=0$ with probability $1$, then $X, Y$ are independent and $Y, Z$ are independent, but $Y$ and $Z$ are arbitrary, and can depend on each other.

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