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Poglavlje 2 Slučajna varijabla

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Numeričke karakteristike slučajne varijable 125<br />

Da bismo lakše riješili potrebne integrale uočimo prvo da je funkcija<br />

g(x) =x 1 x2<br />

− √ e 2<br />

2π<br />

neparna, na R integrabilna funkcija, pa je<br />

�∞<br />

−∞<br />

g(x) dx =0.<br />

Osim toga, zbog normiranosti funkcije gustoće neprekidne slučajne varijable<br />

znamo da je<br />

�∞<br />

−∞<br />

1 x2<br />

− √ e 2 dx =1.<br />

2π<br />

Primjenom supstituciije t =(x − μ)/σ slijedi:<br />

EX =<br />

EX 2 =<br />

=<br />

= μ,<br />

1<br />

√ 2πσ<br />

1<br />

√ 2π<br />

1<br />

√ 2πσ<br />

= σ2<br />

√ 2π<br />

= σ2<br />

√ 2π<br />

�∞<br />

(x−μ)2<br />

−<br />

xe 2σ2 dx = 1<br />

√<br />

2π<br />

�∞<br />

−∞<br />

−∞<br />

�∞<br />

t2<br />

−<br />

σt e 2 dt + μ<br />

−∞<br />

1<br />

�∞<br />

t2<br />

− √ e 2 dt<br />

2π<br />

−∞<br />

�∞<br />

−∞<br />

∞<br />

�<br />

−∞<br />

∞<br />

�<br />

−∞<br />

x 2 (x−μ)2<br />

−<br />

e 2σ2 dx = 1<br />

√<br />

2π<br />

t 2 t2<br />

−<br />

e 2 dt + 2σμ<br />

√<br />

2π<br />

�∞<br />

−∞<br />

t 2 t2<br />

−<br />

e 2 dt + μ 2 = σ 2<br />

�<br />

2<br />

π<br />

�∞<br />

−∞<br />

t2<br />

−<br />

(σt + μ) e 2 dt<br />

(σt + μ) 2 t2<br />

−<br />

e 2 dt<br />

t2<br />

−<br />

te 2 dt + μ2<br />

√<br />

2π<br />

�∞<br />

0<br />

�∞<br />

−∞<br />

t 2 t2<br />

−<br />

e 2 dt + μ 2 .<br />

t2<br />

−<br />

e 2 dt

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