## Download PDF by Edwin P. Alexander: American Locomotives. A Pictorial Record of Steam Power

By Edwin P. Alexander

Американские локомотивы 1900-1950

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**Example text**

8 (Length bound). For X ∼ N (0, 1) and any (Borel) set B, P [X ∈ B] ≤ |B|/2 −|B|/2 |B| 2 fX (x) = 1 − 2Q , where |B| is the length (Lebesgue measure) of the set B. This is because the probability is concentrated around 0. More generally, for X ∼ N (m, σ 2 ) |B| 2σ P [X ∈ B] ≤ 1 − 2Q . 9 (Stein’s Lemma). Let X ∼ N (μ, σ 2 ), and let g be a diﬀerentiable function satisfying E |g (X)| < ∞. Then E [g(X)(X − μ)] = σ 2 E [g (X)] . 5 p 124]. Note that this is simply integration by parts with u = g(x) and dv = (x − μ)fX (x)dx.

E λ ! 15. P [X ≥ 2] = 1 − e−λ − λe−λ = O (λ2 ) . d. E(λ). The equalities given by (*) are easily obtained via counting the number of events from rate-λ Poisson process on interval [0, 1]. 16. Fano factor (index of dispersion): Var X EX =1 An important property of the Poisson and Compound Poisson laws is that their classes are close under convolution (independent summation). 31) which are straightforward to prove from their characteristic functions. 17 (Recursion equations). Suppose X ∼ P(λ).

Starting with m1 = λ = μ2 and μ1 = 0, the above equations lead to recursive determination of the moments mk and μk . 18. E as 1 X+1 1 λ = d−1 n n=0 bn X + 1 − e−λ . Because for d ∈ N, Y = c , X+1 1 X+1 d n=0 an X n can be expressed the value of EY is easy to ﬁnd if we know EX n . 19. Mixed Poisson distribution: Let X be Poisson with mean λ. Suppose, that the mean λ is chosen in accord with a probability distribution whose characteristic function is ϕΛ . Then, ϕX (u) = E E eiuX |Λ = E eΛ(e iu −1 ) = E ei(−i(eiu −1))Λ = ϕ −i eiu − 1 Λ .

### American Locomotives. A Pictorial Record of Steam Power 1900-1950 by Edwin P. Alexander

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