Say we have two random variables, $X$ and $Z$ that are independent. Then let $W=a+bXZ$ be the random variable that is a function of both $X$ and $Z$, $a$ and $b$ are just scalar constants. Then are $X$ and $W$ independent? How about $Z$ and $W$?

This is just something I thought of, not sure how I would go about proving (or disproving) it.


The random variables $W$ and $X$ are called independent when their joint probability $P(W,X)$ equals the product of their marginal probabilities, $P(W)P(X)$, or, in other words, the conditional probability is $P(W|X)\stackrel{\mathrm{def}}{=}P(W,X)/P(X)=P(W)$.

You can establish independence by computing $P(W|X)$ and also $P(W)$, and checking if they are equal. This wikipedia page might help you compute these distributions when some of them are normal.

  • $\begingroup$ Thanks Kirill, so to make things more concrete I have: $X \sim InverseGamma(shape=v/2,scale=v/2)$ and $Z \sim N(0,1)$. Now clearly, $W|X \sim N(a, (bX)^2)$, how can I find the distribution of $W$? $\endgroup$ – Trts Oct 5 '13 at 17:40
  • $\begingroup$ The marginal distribution of $W$ is defined as the integral of the joint distribution over all the other variables: $P(w) = \int P(w,x,z)\,dx\,dz.$ Notice that the distribution $P(w)$ will be a function of $w$ only, not $x$, nor $z$. $\endgroup$ – Kirill Oct 5 '13 at 17:43
  • $\begingroup$ Yup, so I found $\displaystyle P(x,z) \propto \frac{1}{x^{(v+1)}}\exp\left\{-\frac{1}{2}\left[\frac{v}{x^2}+z^2\right]\right\}$ using the independence between $X$ and $Z$, how do I find $P(w,x,z)$? $\endgroup$ – Trts Oct 5 '13 at 17:48
  • $\begingroup$ The distribution of $P(w,x,z)$ is precisely what you gave in the question: $P(w,x,z) = P(x)P(z)\delta(w-(a+bxz))$, i.e., you yourself defined it. In this case, you already said what the distribution of $W|X$ is, so you can just compute $P(w) = \int P(w|x)P(x)\,dx$, and check whether it equals $P(w|x)$. (I think you are making this sound harder than it really is.) $\endgroup$ – Kirill Oct 5 '13 at 17:57
  • $\begingroup$ Ah yes, I certainly was overcomplicating things, thanks Kirill for clearing that up! $\endgroup$ – Trts Oct 5 '13 at 18:08

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