# Moments of a generic random variable (symbolic)

GROUPS:
 Daniela Scida 1 Vote Hi, I'm having trouble to find how write moments of random variables in symbolic terms. For instance, say I define z = x + y then I want to compute the variance of z in terms of moments of x and y. I would like Mathematica to return this:Variance(z) = variance(x) + variance(y) + 2*covariance(x,y)(All in symbolic terms) My actual problem is a bit more complicated. I have that the data generating process of a scalar random variable X is an AR(p): $X_{t} = \sum_{j=1}^{p} A_{j} X_{t-j} + u_{t}$where $A_{j}$'s are scalars, for some white noise $u_{t}$. I don't want to specify a distribution for $u_{t}$, and there is NO data nor estimation involved. This is a population exercise. I want to compute the variance of $X_{t}$ as a function of $p$, the $A_{j}$'s, and probably the variance of $u_{t}$. The process is stationary.MY QUESTION: can mathematica give me the expression of Variance( $X_{t}$ ) in symbolic terms? Thanks!!!! All help/comments welcome!PS Just in case I'm familiar with the well known formula for it, but I don't want to write that directly. I want mathematic to tell me the expression. The reason is that I also have more complicated processes to look into where deriving the variance is a mess, so I would like mathematica to do it for me. For instance, I have VAR(p) processes for X and Y.
4 years ago
8 Replies
 Jim Baldwin 2 Votes You might consider the mathStatica book/package at mathstatica.comAlso, you have a minor typo: "- 2 * covariance(x,y)" should be "+ 2 * covariance(x,y)".
4 years ago
 @Jim Thanks! I actually looked into mathStatica before, but it seems you need to buy it? (And yes there was a minor, I was tired when I wrote it...Thanks!)
4 years ago
 Yes, one does need to purchase mathStatica but it is well worth it.
4 years ago
 Awesome! Thanks! I'll keep that in mind.
4 years ago
 Gosia Konwerska 4 Votes Assuming weak stationarity of AR process: In[1]:= Variance[ARProcess[{a1, a2}, var][t]] Out[1]= -(((-1 + a2) var)/((1 + a2) (1 - a1^2 - 2 a2 + a2^2))) gives the answer for p=2 and for var denoting the variance of white noise driving the process. In Mathematica ARProcess is assumed to be driven by a Gaussian white noise, but this formula is more general - assuming zero first moment and finite second moment of the noise. We also assume that the noise variable is independent of past observations.Clearly it becomes more complicated in higher dimensions: In[2]:= Variance[ARProcess[{Array[a, {2, 2}]}, Array[v, {2, 2}]][t]] Out[2]= {-((v[1, 1] - a[1, 2] a[2, 1] v[1, 1] - a[1, 1] a[2, 2] v[1, 1] - a[2, 2]^2 v[1, 1] - a[1, 2] a[2, 1] a[2, 2]^2 v[1, 1] + a[1, 1] a[2, 2]^3 v[1, 1] + a[1, 1] a[1, 2] v[1, 2] + a[1, 2]^2 a[2, 1] a[2, 2] v[1, 2] - a[1, 1] a[1, 2] a[2, 2]^2 v[1, 2] + a[1, 1] a[1, 2] v[2, 1] + a[1, 2]^2 a[2, 1] a[2, 2] v[2, 1] - a[1, 1] a[1, 2] a[2, 2]^2 v[2, 1] + a[1, 2]^2 v[2, 2] - a[1, 2]^3 a[2, 1] v[2, 2] + a[1, 1] a[1, 2]^2 a[2, 2] v[2, 2])/(-1 + a[1, 1]^2 + a[1, 2] a[2, 1] + a[1, 1]^2 a[1, 2] a[2, 1] + a[1, 2]^2 a[2, 1]^2 - a[1, 2]^3 a[2, 1]^3 + a[1, 1] a[2, 2] - a[1, 1]^3 a[2, 2] + 3 a[1, 1] a[1, 2]^2 a[2, 1]^2 a[2, 2] + a[2, 2]^2 - a[1, 1]^2 a[2, 2]^2 + a[1, 2] a[2, 1] a[2, 2]^2 - 3 a[1, 1]^2 a[1, 2] a[2, 1] a[2, 2]^2 - a[1, 1] a[2, 2]^3 + a[1, 1]^3 a[2, 2]^3)), -((-a[2, 1]^2 v[1, 1] + a[1, 2] a[2, 1]^3 v[1, 1] - a[1, 1] a[2, 1]^2 a[2, 2] v[1, 1] - a[1, 1] a[1, 2] a[2, 1]^2 v[1, 2] - a[2, 1] a[2, 2] v[1, 2] + a[1, 1]^2 a[2, 1] a[2, 2] v[1, 2] - a[1, 1] a[1, 2] a[2, 1]^2 v[2, 1] - a[2, 1] a[2, 2] v[2, 1] + a[1, 1]^2 a[2, 1] a[2, 2] v[2, 1] - v[2, 2] + a[1, 1]^2 v[2, 2] + a[1, 2] a[2, 1] v[2, 2] + a[1, 1]^2 a[1, 2] a[2, 1] v[2, 2] + a[1, 1] a[2, 2] v[2, 2] - a[1, 1]^3 a[2, 2] v[2, 2])/(1 - a[1, 1]^2 - a[1, 2] a[2, 1] - a[1, 1]^2 a[1, 2] a[2, 1] - a[1, 2]^2 a[2, 1]^2 + a[1, 2]^3 a[2, 1]^3 - a[1, 1] a[2, 2] + a[1, 1]^3 a[2, 2] - 3 a[1, 1] a[1, 2]^2 a[2, 1]^2 a[2, 2] - a[2, 2]^2 + a[1, 1]^2 a[2, 2]^2 - a[1, 2] a[2, 1] a[2, 2]^2 + 3 a[1, 1]^2 a[1, 2] a[2, 1] a[2, 2]^2 + a[1, 1] a[2, 2]^3 - a[1, 1]^3 a[2, 2]^3))} 
 Gosia Konwerska 2 Votes In[1]:= ARProcess[{Array[a, {2, 2}]}, Array[v, {2, 2}]] Out[1]= ARProcess[{{{a[1, 1], a[1, 2]}, {a[2, 1], a[2, 2]}}}, {{v[1, 1], v[1, 2]}, {v[2, 1], v[2, 2]}}] denotes AR process of order p=1, dimension 2. The coefficients of a vector process are defined to be square matrices n by n, where n is the dimension of the process. So to define 2-dimensional AR process of order say p=4 with symbolic coefficients using Array one does: In[2]:= ARProcess[{Array[a, {4, 2, 2}]}, Array[v, {2, 2}]] Out[2]= ARProcess[{{{{a[1, 1, 1], a[1, 1, 2]}, {a[1, 2, 1], a[1, 2, 2]}}, {{a[2, 1, 1], a[2, 1, 2]}, {a[2, 2, 1], a[2, 2, 2]}}, {{a[3, 1, 1], a[3, 1, 2]}, {a[3, 2, 1], a[3, 2, 2]}}, {{a[4, 1, 1], a[4, 1, 2]}, {a[4, 2, 1], a[4, 2, 2]}}}}, {{v[1, 1], v[1, 2]}, {v[2, 1], v[2, 2]}}] One could also use different names for each coefficient matrix: In[3]:= ARProcess[{Array[a, {3, 3}], Array[b, {3, 3}]}, Array[v, {3, 3}]] Out[3]= ARProcess[{{{a[1, 1], a[1, 2], a[1, 3]}, {a[2, 1], a[2, 2], a[2, 3]}, {a[3, 1], a[3, 2], a[3, 3]}}, {{b[1, 1], b[1, 2], b[1, 3]}, {b[2, 1], b[2, 2], b[2, 3]}, {b[3, 1], b[3, 2], b[3, 3]}}}, {{v[1, 1], v[1, 2], v[1, 3]}, {v[2, 1], v[2, 2], v[2, 3]}, {v[3, 1], v[3, 2], v[3, 3]}}] which defines AR process of order 2 and dimension 3, with coefficient matrices a and b, and noise covariance matrix v.This may also be helpful: ARProcess reference page