# Independence in a sum relationship

I have a model of total reaction time T, which is a composite of a selection time S and a discrimination time D. So a person first finds something, this takes the tS. Then he discriminates and reports an attribute of this thing, this takes the time D. We only measure the total time T, though. So, my model for T is:

T = S + D

In considering, I try to think about how these random variables behave in simplified settings; and these are my questions:

1.) If I would define S and D to be independent, could I infer that S and T, or D and T are independent? If not, could I infer that D and T are conditionally independent, given S?

2.) Defining S and D to be independent and normally distributed, I know I can infer that T is normally distributed. But what if I define T and D as normally distributed, can I infer that S is normally distributed? I think this boils down to the question 1.) whether I can say that T and D are independent.

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T = S+D

1.) If I would define S and D to be independent, could I infer that S and T, or D and T are independent? If not, could I infer that D and T are conditionally independent, given S?

No, $S$ and $T$ are not independent except in the trivial case when $S$ has variance $0$, that is, $S$ is a degenerate random variable that takes on fixed value $\mu_S$ with probability $1$. In nontrivial cases, $S$ and $T$ are positively correlated. This follows from the fact that covariance is a bilinear function. We have that

$$\text{cov}(S, T) = \text{cov}(S, S+D) = \text{cov}(S, S)+\text{cov}(S, D) = \text{var}(S) + 0 \geq 0.$$

Correlated random variables cannot be independent. Similarly, $D$ and $T$ are positively correlated and thus cannot be independent.

Nor can $D$ and $T$ be conditionally independent given the value of $S$. Remember that given $S = \alpha$, $T = S+D = \alpha+D$ differs from $D$ only by the constant $\alpha$. How can $T$ and $D$ be (conditionally) independent in view of the fact that knowing the value of one allows us to predict the value of the other with perfect accuracy and certainty?

2.) Defining S and D to be independent and normally distributed, I know I can infer that T is normally distributed. But what if I define T and D as normally distributed, can I infer that S is normally distributed? I think this boils down to the question 1.) whether I can say that T and D are independent.

As Michael Chernick has pointed out, if $T$ and $D$ are normal, their difference $S$ is not necessarily normal; joint normalcy of $T$ and $D$ (including as a special case $T$ and $D$ being independent normals) is required. Another condition is as follows.

If $T = S+D$ is normal and $S$ and $D$ are independent, then $S$ and $D$ also are normal random variables. See here about this decomposition. for more information

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 +1 You managed to improve enough on my answer for the OP to give you the check mark.@DilipSarwate. – Michael Chernick Oct 9 '12 at 17:03 @MichaelChernick Thanks for the up vote. – Dilip Sarwate Oct 9 '12 at 17:05

The answer to 1 is no. It does matter what you assume about how S and D are related, T is a function of both S and D. So it is correlated (positively) with both S and D and given S, D and T are still positively correlated.

I think the answer to 2 is no also because T and D could be marginally normal without being bivariate normal and then you cannot assert that all linear combinations of T and D are normal.

On the other hand in 2 if you assume T and D are bivariate normal then since S=T-D, S is normal.

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