Recall that the Fourier transform (also known as characteristic function) of a random variable always exists, given by. e itx f(x)dx.

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1 I am going to talk about algebraic calculations of random variables: i.e. how to add, subtract, multiply, divide random variables. A main disadvantage is that complex analysis is used often, but I will skip those part and focus on others. A previous post called Generating Functions and Transforms is assumed. From now on, assume the random variables X, X 2,... are absolutely continuous (have pdf s) and independent. For the discrete case, we use brute force and it s not interesting. This post relies heavily on [], but I find it really useful and maybe I will come back to make more examples. Recall that the Fourier transform (also known as characteristic function) of a random variable always exists, given by F X (t) = E(e itx ) = e itx f(x)dx and also F X+Y (t) = F X (t)f Y (t). In most cases, the inversion formula is given by or equivalently, f(x) = 2π F (x) = F () + 2π e itx F X (t)dt We use the above facts to reach the following formulas: e itx F X (t)dt it Theorem. (+,-) () Define U = n j= X j, then the pdf of U is given by g(u) = 2π e itu n j= (2) Define V = X X 2 then the pdf of V is given by g(v) = 2π F Xj (t)dt e itv F X (t)f X2 (t)dt You may think that why don t we use the convolution method given by f X +X 2 (x) = f (x x 2 )f 2 (x 2 )dx 2 this is applicable but more comprehensive when you have many random variables to add. We can also use Laplace transform, but we have to assume those random variables are all nonnegative. Then, the pdf of U is given by g(u) = i i e ru n j= L Xj (r)dr

2 2 Example. (sum of two rv s) Let X Uniform([,]); X 2 Exponential(). W := X + X 2. We calculate the Laplace transform, we have L X (r) = L X2 (r) = e rx dx = e r r e rx e x dx = r + (note that the upper limit in the first integral is ). Therefore, g(w) = i rw e r e i r(r + ) dr I should have mentioned earlier that the integral is calculated through the residue theorem in complex analysis. The integral above is a line integral, where you integrate in the complex plane along a line, but we don t calculate it directly, we close the line using another curve to get a contour integral. We can prove that the integral of the curve is, using Jordan s lemma which is proved by estimation theorem of complex integrals. To evaluate the contour integral, we apply residue theorem, which simplifies calculating the integral into calculating residues, which is essentially the coefficient of (z a) in the Laurent expansion of the integrand at an undefined point a. Back to the example, we have g(w) = g (w) g 2 (w) where g (w) = i e rw i r(r + ) dr g 2 (w) = i e r(w ) i r(r + ) d Jordan e lemma holds when w > for g, and w > for g 2, elsewhere they are. By residue s theorem, g (w) = erw ()[( r + ) r= + ( erw r ) r= ] = e w Hence we have g 2 (w) = ()[(er(w ) r + ) r= + ( er(w ) ) r= ] = e w r, if w g(w) = e w, if < w e w e w, if w > () Now we look at product and quotient. First we assume X, X 2,... to be nonnegative, and

3 3 later for the general case. Recall that Mellin transform is given by and its inversion also M XY (s) = M X (s)m Y (s). M X (s) = f(x) = lim T c+t i c T i x s f(x)dx x s M X (s))ds The first method is similar to the convolution you know before. By a few manipulations through Jacobian and change of variables we have that, the distrbution of U = X X 2 is given by h(u) = f ( u )f 2 (x 2 )dx 2 = x 2 x 2 which is called the Mellin convolution. Similarly, if V = X X 2 h(v) = x 2 f (vx 2 )f 2 (x 2 )dx 2 x f (x )f 2 ( u x )dx then its distribution is given by However, they may be too complicated for many random variables, hence we have the following: Theorem. (, ) () Define U = n j= X j, then the pdf of U is given by g(u) = n u s j= (2) Define V = X /X 2 then the pdf of V is given by g(v) = M Xj (s)ds v s M X (s)m X2 (2 s)ds Division is not that intuitive as subtraction. We consider Y = X a for general a, M Y (s) = E(y s ) = Letting a = yields the desired result. x a(s ) f(x)dx = M X (as a + ) However, in practice the calculation is very involved as you may see in []. I only give two relatively simple examples:

4 4 Example. (, ) Let X,..., X n Uniform([,]) be independent. () Let U := n j= X j, then the Mellin transform of X i is hence by formula, We apply Jordan s lemma, M Xi (s) = x s dx = s g(u) = u s s n ds g(u) = (2) Let V := X X 2, by formula we have (n )! [ dn ds n (u s )] s= = (ln( u )n ) (n )! g(v) = v s s(2 s) ds now we have to separate cases according to Jordan e lemma, and different contours are involved! I will directly give you the result: For v, For v >, g(v) = v s ()[ 2 s ] s= = 2 g(v) = ()[v s s ] s=2 = 2v 2 Now, what about random variables taking negative values? Our trick is to separate the random variable into two parts, say for a distribution function f(x) we define { f +, if x < (x) = (2) f(x), if x f (ω) = {, if x f(x), if x < Naturally we have f = f + + f. (If you read the Lebesgue integral part you should note that this is different from the so called positive and negative parts ). Now consider the product XY, if the product is positive we have two cases: X >, Y > ; X <, Y <, similar for negative. Without loss of generality we discuss the case XY is positive, recall Mellin convolution h(u) = f (x )f 2 ( u )dx = (f + x x x (x ) + f (x ))(f 2 + ( u ) + f2 x ( u ))dx x (3)

5 5 consider the two cases we have, the correspond to the following h(u) = and now we can use Mellin s transform. x (f + (x )f + 2 ( u x ) + f ( x )f 2 ( u x ))dx The last task is when we have both addition and multiplication of random variables. Since we use different techniques when adding and multiplying, we must transform back and forth between different transforms. Theorem. (transforming between transformations) Under certain circumstances, M X (s) = Γ(s) L X (r) = F X (t) = L X (r)( r) s dr M X (s)γ( s)r s ds M X (s)γ( s)(it) s ds Bonus In [], the author introduced H-function, but I doubt if we can understand it in practice. H-function is a generalized form of distributions, which makes it convenient to multiply random variables, because there is a formula for multiplying H-function random variables. However, the proof of the formula requires complex analysis and I skip it. Definition. (H-Function) The H-function is given by H m,n p,q [z (a, α ),..., (a p, α p ) (b, β ),..., (b q, β q ) ] =: H(z) = c+i m j= Γ(b j β j s) n j= Γ( a j + α j s) q j=m+ Γ( b j + β j s) p j=n+ Γ(a j α j s) zs ds where: m q; n p; α j >, β j > ; a j, b j C The H-function distribution is given by: { kh(cx) if x f(x) = otherwise (4)

6 6 The characteristic function of the H-distribution is given by ϕ(t) = k c Hn+,m q,p+ [ it c ( b β, β ),..., ( b q β q, β q ) (, ), ( a α, α ),..., ( a p α p, α p ) ] And the moments can be derived by the following formula: m n = k m j= Γ(b j + β j + β j r) n j= Γ( a j α j α j r) c r+ q j=m+ Γ( b j β j β j r) p j=n+ Γ(a j + α j + α j r) zs ds Common continuous distributions where X > can be represented as H-distributions. For example, the exponential distribution with parameter λ is λh,, [λx (, )] Theorem. (product of H-function rvs) The product of independent random variables with distributions k j H m j,n j p j,q j [c j x j (a j, α j ),..., (a jp, α jp ) ] if x j f j (x j ) = (b j, β j ),..., (b jq, β jq ) (5) otherwise has distribution ( n h(y) = j= k mj, n j j)h pj, q j [ n j= c jy (a, α ),..., (a npn, α npn ) (b, β ),..., (b nqn, β nqn ) ] if y otherwise (6) References: [] Springer, M. D. (979). The algebra of random variables. New York, NY: Wiley.