八 4.Poisson 分配與指數分配

八 4.Poisson 分配與指數分配

4.1 (1)∑_∞

k=0P (k, T ) =∑_∞

k=0 (λT )^k

k! e^{−(λT )} = e^{−(λT )}∑_∞

k=0 (λT )^k

k! = e^{−(λT )}· e^{(λT )} = 1.

(2) E(X) =∑_∞

k=0k·PX(k) =∑_∞

k=0k·^m_k!^ke^−m =∑_∞

k=1 m^k

(k−1)!e^−m = m·e^−m∑_∞

k=1 m^k−1 (k−1)!

= m· e^−m· e^m = m.

E(X²) =∑_∞

k=0k²· PX(k) =∑_∞

k=0k²· ^m_k!^ke^−m =∑_∞

k=0k· _(k^m_−1)!^k e^−m

=∑

[(k− 1) + 1] ·_(k^m_−1)!^k e^−m =∑

(_(k^m_−2)!^k e^−m+ _(k^m_−1)!^k e^−m)

= m²e^−m∑_∞

k=2 m^k−2

(k−2)! + me^−m∑_∞

k=1 m^k−1

(k−1)! = m²+ m.

V ar(X) = E(X²)− E(X)² = m²+ m− m² = m.

4.2 P_Y(k) =∑_k

i=1P_X1(i)P_X2(k− i) =∑_k

i=1(^m_i!ⁱe^−m)(_(k−i)!^mk−ie^−m) =∑_k

i=1

i!(k−i)!1 m^ke^−2m

=∑k i=1

1 k!

k!

i!(k−i)!m^ke^−2m = _k!¹m^ke^−2m∑_k

i=1C_i^k = _k!¹m^ke^−2m2^k= ^(2m)_k!^ke^−2m.

4.3 P (k) = ^m_k!^ke^−m = ^m₁^·m···m_·2···k e^−m= ^m₁ ·^m₂ ·^m₃ · · · ·^m_ke^−m, k為正整數, 當k = m − 1, m時, P (k) = ₁·2···(m−1)^m·m···m e^−m= ^m₁ · ^m₂ ·^m₃ · · · ·_m^m₋₁e^−m 為最大值.

4.4 PZ(k) =∑k

i=1PX(i)PY(k− i) = ∑k

i=1(^m_i!ⁱe^−m)(_(k^lk−i_−i)!e^−l) =∑k

i=1(^m_i!(k^i·l_−i)!^k−ie^−m−l)

= e^−m−l· _k!¹ ∑_k

i=1 k!

i!(k−i)!mⁱl^k−i = e^−(m+l)·^(m+l)_k! ^k 4.5 由4.4可得 PZ(k) = e^−(nm)· ^(nm)_k! ^k

4.6 f_Z(t) =∫_t

0 λe^−λ(s−t)· λe^−λ(−s)ds =∫_t

0λ²e^−λtds = λ²te^−λt.

4.8 (a) 每頁平均可以發現1個錯誤, 所以可設機率函數為P (k, 1) = ^(1)k_k!^·e−1 = ^e_k!⁻¹. 因此, 在某頁發現1個錯誤的機率為P (1, 1) = e⁻¹ = 0.368.

(b) 可設在4頁中發現錯誤的機率函數為P (k, 4) = ^4k·e_k!⁻⁴. 因此, 在4頁發現4個錯誤的機率 為P (4, 4) = ^44·e_4!⁻⁴ = 0.195

4.9 平均每天發生¹⁰₃次車禍, 所以可設機率函數為P (k,¹⁰₃) = ⁽

10 3)^k·e^{− 103}

k! . 因此,10天內僅發生一件車禍的機率為P (1,¹⁰₃ ) = (¹⁰₃)· e⁻¹⁰³ = 0.119.

4.11 串聯之可靠度為e^−ntµ, E( ˆL) = ∫_∞

0 e^−ntµdt =∫_∞

0 e^−ntµdt =−^µ_ne^−ntµ|^∞₀ = ^µ_n. 並聯之可靠度為1 − (1 − e^−µt)ⁿ =∑n

k=1C_kⁿ(−1)^k+1e^−ktµ, E( ˆL) =∫_∞

0

∑n

k=1C_kⁿ(−1)^k+1e^−ktµdt =∑n

k=1C_kⁿ(−1)^k+1∫_∞

0 e^−ktµdt =∑n

k=1C_kⁿ(−1)^{k+1 µ}_k

= µ∑n

k=1C_kⁿ(−1)^{k+1 1}_k = (1 + ¹₂ +¹₃ +· · · +_n¹)µ.

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