By Ronald Meester
Compactly written, yet however very readable, beautiful to instinct, this advent to likelihood conception is a superb textbook for a one-semester direction for undergraduates in any path that makes use of probabilistic rules. Technical equipment is just brought while precious. The course is rigorous yet doesn't use degree concept. The textual content is illustrated with many unique and fantastic examples and difficulties taken from classical purposes like playing, geometry or graph idea, in addition to from purposes in biology, medication, social sciences, activities, and coding concept. merely first-year calculus is needed.
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Additional resources for A Natural Introduction to Probability Theory, Second Edition
To compute its expectation, we write ∞ k(1 − p)k−1 . E(X) = p k=1 n ∞ Let us denote k=1 k(1 − p)k−1 by Sn , and k=1 k(1 − p)k−1 by S. We would like to compute S, and there are various ways of doing this. 36 for an alternative approach. The point is that we ∞ recognise S as the derivative (with respect to p) of − k=0 (1 − p)k , and this last 52 Chapter 2. Random Variables and Random Vectors expression is equal to −1/p. It follows that S = p−2 . 6(b). 36. Here is an alternative way of computing S, without using diﬀerentiation.
Proof. First, E(Y ) = jP (Y = j) = j j j P (X = i, Y = j) i jP (X = i, Y = j), = i j and therefore, iP (X = i) + E(X) + E(Y ) = i = jP (X = i, Y = j) i i j P (X = i, Y = j) + i j = jP (X = i, Y = j) i j (i + j)P (X = i, Y = j) i j z j zP (X = z − j, Y = j) = = z z = P (X + Y = z, Y = j) j zP (X + Y = z). z It follows that the last sum is well deﬁned and hence E(X + Y ) exists and is equal to E(X) + E(Y ). We make two remarks concerning the expectation of sums. 1. The conditions of the theorem are not necessary for the existence of E(X+Y ).
16. 13. 17. Suppose that X has distribution function F . What is the distribution function of the random variable Y deﬁned by Y = aX + b? We end this section with a statement to the eﬀect that the probability mass function of a random variable is uniquely determined by its distribution function, and vice versa. 18. Two random variables have the same probability mass function if and only if they have the same distribution function. Proof. Let X and Y be such that pX (x) = pY (x) for all x. Then FX (x) = P (X ≤ x) = pX (y) y:y≤x pY (y) = FY (x).
A Natural Introduction to Probability Theory, Second Edition by Ronald Meester