I IID and Dependent RVs: A Closer Look at their Relationship and Parameters

[member 428835]

If $\epsilon_1,\epsilon_2$ are iid $N(0,1)$, $X_1=\mu_1+\sigma_1 \epsilon_1$ and $X_2=\mu_2+\rho\epsilon_1+\sigma_2 \epsilon_2$ are evidently a pair of dependent RVs that are not identically distributed for most values of the parameters. I have no idea what $\mu,\sigma,\rho$ are. I assume $\mu$ is mean and $\sigma$ is standard deviation? I read this example here.

 

[FactChecker]

If $\epsilon_1,\epsilon_2$ are iid $N(0,1)$, $X_1=\mu_1+\sigma_1 \epsilon_1$ and $X_2=\mu_2+\rho\epsilon_1+\sigma_2 \epsilon_2$ are evidently a pair of dependent RVs that are not identically distributed for most values of the parameters. I have no idea what $\mu,\sigma,\rho$ are. I assume $\mu$ is mean and $\sigma$ is standard deviation? I read this example here.

The example in the link does not say exactly what they are, but we can make the reasonable assumption that he is using the very common notations, where all of the $\mu$s, $\epsilon$s, and $\sigma$s are real number constants EDIT: with $\epsilon$s, and $\sigma$s positive. In that case, $\mu_1$ is mean and $\sigma_1$ is standard deviation of $X_1$. Also, $\mu_2$ is mean of $X_2$. The standard deviation of $X_2$ is more complicated. The SD of the individual terms is $\rho$ and $\sigma_2$, but the sum of those has a variance which is the sum of the variances

 

[member 428835]

The example in the link does not say exactly what they are, but we can make the reasonable assumption that he is using the very common notations, where all of the $\mu$s, $\epsilon$s, and $\sigma$s are real number constants. In that case, $\mu_1$ is mean and $\sigma_1$ is standard deviation of $X_1$. Also, $\mu_2$ is mean of $X_2$. The standard deviation of $X_2$ is more complicated. The SD of the individual terms is $\rho$ and $\sigma_2$, but the sum of those has a variance which is the sum of the variances

Okay, thanks. So I'm missing the crux: why are these dependent instead of independent? It seems to be because $\epsilon_1$ is a function of $X_1$, and so $X_2$ implicitly depends on $X_1$?

 

[FactChecker]

No. $\epsilon_1$ is not a function of $X_1$. It is the other way around.
It is not that $X_2$ depends on $X_1$. It is better to understand that they both depend on $\epsilon_1$, so their tendencies are related. That makes them correlated and not independent.

(PS. I don't like to say that $X_1$ and $X_2$ are dependent until you are comfortable with what that means in probability. It just means that the tendencies of one give a hint to the tendencies of the other. It does not mean the functional dependency that you are probably used to.)

 

[member 428835]

No. $\epsilon_1$ is not a function of $X_1$. It is the other way around.
It is not that $X_2$ depends on $X_1$. It is better to understand that they both depend on $\epsilon_1$, so their tendencies are related. That makes them correlated and not independent.

(PS. I don't like to say that $X_1$ and $X_2$ are dependent until you are comfortable with what that means in probability. It just means that the tendencies of one give a hint to the tendencies of the other. It does not mean the functional dependency that you are probably used to.)

Perfect explanation, thanks so much!

 

[FactChecker]

Consider human arm length and leg length. Clearly, they are related and not independent, yet there are many other factors involved and one does not cause the other. It is just that their tendencies are related.

 

[BWV]

For one concrete example, one can see a similar form in Moving Average models in time series analysis

https://en.wikipedia.org/wiki/Moving-average_model