# Dirichlet forms and markov process by Masatoshi Fukushima

By Masatoshi Fukushima

This publication is an try and unify those theories. by means of unification the speculation of Markov technique bears an intrinsic analytical device of significant use, whereas the speculation of Dirichlet areas acquires a deep probabilistic constitution.

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**Example text**

M. 2 Discrete and Continuous Random Variables −1 ???? e 2 =√ 2???? ( )2 ???? ???? ( )2 23 ( )2 [ ( ???? )] [ ( ???? )] ???? −1 ???? +√ e 2 ???? −???? 1−Φ − −???? 1−Φ ???? ???? 2???? ( ???? )] [ (????) −Φ − +???? Φ ???? ???? 2???? − 12 ???????? =√ e 2???? √ ( )2 [ ( ???? )] −1 ???? 2 =???? . e 2 ???? + ???? 1 − 2Φ − ???? ???? ( ) To evaluate ???? Y 2 by definition, ( ) ???? Y2 = ∞ ∫−∞ ∞ = ∫0 y2 fY (y) dy y2 −1 e 2 √ ???? 2???? ( ) y+???? 2 ???? ∞ dy + ∫0 y2 −1 e 2 √ ???? 2???? ( ) y−???? 2 ???? dy. By setting z = (y + ????)∕???? and ???? = (y − ????)∕???? we have ( ) ???? Y2 = ∞ ∫????∕???? ∞ 1 1 1 1 2 − z2 2 − ????2 √ (????z − ????) e 2 dz + √ (???????? + ????) e 2 d???? ∫−????∕???? 2???? 2???? ∞ ∞ ∞ 1 2 1 2 1 2 2???????? 1 ????2 =√ z2 e− 2 z dz − √ ze− 2 z dz + ????2 e− 2 z dz √ ∫ ∫ ∫ ????∕???? 2???? ????∕???? 2???? ????∕???? 2???? ∞ ∞ 1 2 1 2 2???????? ????2 +√ ????2 e− 2 ???? d???? + √ ????e− 2 ???? d???? ∫ ∫ 2???? −????∕???? 2???? −????∕???? ∞ + ????2 1 2 1 e− 2 ???? d???? √ 2???? ∫−????∕???? ] [ ( )2 ( ???? )) ( ???? ) 1 ( ???? )2 √ ( 2???????? − 12 ???????? −2 ???? ????2 + 2???? 1 − Φ e −√ =√ e ???? ???? 2???? 2???? [ ( ???? )] + ????2 1 − Φ ???? ] [ ( )2 ( )2 ( ( ???? )) √ ( 2???????? − 12 ???????? ???? ) − 12 ???????? ????2 e +√ e − +√ + 2???? 1 − Φ − ???? ???? 2???? 2???? [ ( ???? )] + ????2 1 − Φ − ???? ( ???? )] (????) ( 2 )[ 2 = ???? +???? −Φ − 2−Φ ???? ???? = ????2 + ???? 2.

M. 3 Properties of Expectations holds. Finally, if X and Y are a pair of jointly continuous variables, show that ???? (|XY|) ≤ {???? (|X p |)}1∕p {???? (|Y q |)}1∕q . Solution: The inequality certainly holds for ???? = 0 and ???? = 0. Let ???? = ex∕p and ???? = ey∕q where x, y ∈ ℝ. By substituting ???? = 1∕p and 1 − ???? = 1∕q, e????x+(1−????)y ≤ ????ex + (1 − ????)ey holds true since the exponential function is a convex function and hence ???????? ≤ By setting ????= hence ????p ???? q + . p q |X| , {???? (|X p |)}1∕p ????= |Y| {???? (|Y q |)}1∕q |X|p |Y|q |XY| ≤ + .

J∈N \ S ( ) n combinations to select sets of indices i1 , . . , ik from N, which k are mutually exclusive events, so However, as there are ℙ(X = k) = ( ) n k p (1 − p)n−k , k k = 0, 1, 2, . . , n. From the definition of the moment generating function of discrete random variables (see Appendix B), ( ) ∑ tx MX (t) = ???? etX = e ℙ(X = x) x ( ) n x where t ∈ ℝ and substituting ℙ(X = x) = p (1 − p)n−x we have x MX (t) = n ∑ x=0 ( ) n ( ) ∑ n x n n−x p (1 − p) (pet )x (1 − p)n−x = (1 − p + pet )n . M.