Answer to Question #85925 in Statistics and Probability for Umalaxmi
2019-03-06T09:40:06-05:00
For the random variable X with density function f(x)={2e^(-2x) ; x≥0
{0 ;x<0.
Find (i)the mean μ and variance σ^(2);
(ii)P [|X-μ|≥1].
Use Chebyshev's inequality to obtain an upper bound on P[|X-μ|]≥1] and compare with the result obtained in part (ii).
1
2019-03-08T08:15:33-0500
μ = ∫ R x f ( x ) d x = ∫ 0 ∞ 2 x e − 2 x d x = 0.5 \mu=\int\limits_{\mathbb{R}} xf(x) \, dx=\int\limits_0^{\infty} 2xe^{-2x}\,dx=0.5 μ = R ∫ x f ( x ) d x = 0 ∫ ∞ 2 x e − 2 x d x = 0.5
σ 2 = ∫ R ( x − μ ) 2 f ( x ) d x = ∫ 0 ∞ 2 ( x − 0.5 ) 2 e − 2 x d x = 0.25 \sigma^2=\int\limits_{\mathbb{R}} (x-\mu)^2 f(x) \, dx=\int\limits_0^{\infty} 2(x-0.5)^2 e^{-2x}\,dx=0.25 σ 2 = R ∫ ( x − μ ) 2 f ( x ) d x = 0 ∫ ∞ 2 ( x − 0.5 ) 2 e − 2 x d x = 0.25
P ( ∣ X − μ ∣ ⩾ 1 = 2 σ ) = ∫ ∣ x − 0.5 ∣ ⩾ 1 f ( x ) d x = ∫ 1.5 ∞ 2 e − 2 x d x = e − 3 P(|X-\mu| \geqslant 1=2 \sigma)=\int\limits_{|x-0.5|\geqslant 1} f(x) \, dx=\int\limits_{1.5}^{\infty} 2e^{-2x}\,dx=e^{-3} P ( ∣ X − μ ∣ ⩾ 1 = 2 σ ) = ∣ x − 0.5∣ ⩾ 1 ∫ f ( x ) d x = 1.5 ∫ ∞ 2 e − 2 x d x = e − 3
P ( ∣ X − μ ∣ ⩾ 1 = 2 σ ) ⩽ 1 2 2 = 0.25 P(|X-\mu| \geqslant 1=2 \sigma) \leqslant \dfrac{1}{2^2}=0.25 P ( ∣ X − μ ∣ ⩾ 1 = 2 σ ) ⩽ 2 2 1 = 0.25
e − 3 ≈ 0.05 < 0.25 e^{-3}\approx 0.05<0.25 e − 3 ≈ 0.05 < 0.25
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