<P> f Z (z) = ∫ − ∞ ∞ ∫ − ∞ ∞ f X (x) f Y (y) δ (z − x y) d y d x = ∫ − ∞ ∞ f X (x) f Y (z / x) (∫ − ∞ ∞ δ (z − x y) d y) d x = ∫ − ∞ ∞ f X (x) f Y (z / x) 1 x d x . (\ displaystyle (\ begin (array) (lcl) f_ (Z) (z) & = & \ int _ (- \ infty) ^ (\ infty) \ int _ (- \ infty) ^ (\ infty) f_ (X) \ left (x \ right) f_ (Y) \ left (y \ right) \ delta \ left (z - xy \ right) \, dy \, dx \ \ & = & \ int _ (- \ infty) ^ (\ infty) f_ (X) \ left (x \ right) f_ (Y) \ left (z / x \ right) \ left (\ int _ (- \ infty) ^ (\ infty) \ delta \ left (z - xy \ right) \, dy \ right) \, dx \ \ & = & \ int _ (- \ infty) ^ (\ infty) f_ (X) \ left (x \ right) f_ (Y) \ left (z / x \ right) (\ frac (1) (x)) \, dx. \ end (array))) </P> <P> where we utilize the translation and scaling properties of the Dirac delta function δ (.) (\ displaystyle \ delta \ left (. \ right)). </P> <P> A more intuitive description of the procedure is illustrated in the figure below . The joint pdf f X (x) f Y (y) (\ displaystyle f_ (X) (x) f_ (Y) (y)) exists in the x-y plane and an arc of constant z value is shown as the shaded line . To find the marginal probability f (z) (\ displaystyle f (z)) on this arc, integrate over increments of area d x d y f (x, y) (\ displaystyle dxdy \; f (x, y)) on this contour . </P> <P> We have d y = − d x y / x 2 (\ displaystyle dy = - dx \; y / x ^ (2)) so the increment of probability is δ p = f (x, y) d x d y = f X (x) f Y (z / x) y x 2 (d x) 2 (\ displaystyle \ delta p = f (x, y) dx dy = f_ (X) (x) f_ (Y) (z / x) (\ frac (y) (x ^ (2))) (dx) ^ (2)). If y x d x (\ displaystyle (\ frac (y) (x)) \; dx) can be equated with d z (\ displaystyle dz), then integration over x, yields ∫ f X (x) f Y (z / x) 1 x d x (\ displaystyle \ int f_ (X) (x) f_ (Y) (z / x) (\ frac (1) (x)) dx) yields the integral above . </P>

Expectation of product of two normal random variables