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Problem 1 For Gaussian distribution ρ (x)-ae Find: (1) Constant a; (2) <x> , <x> and standard deviation of the distribution; (3) Sketch the graph p(x) (x-b)2 -T2

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Answer #1

(i)

p(x)dx1

e 2c

cV2

dx=csqrt {2}dt

acsqrt {2}int_{-infty}^{infty}e^{-t^2}dx=1

2acV2e ete 0

V TT

mathbf{a=rac {1}{csqrt {2pi}}}

(ii)

le 2,2 do

=int_{-infty}^{infty}(x-b+b)e^{rac {-(x-b)^2}{2c^2}}dx

=aint_{-infty}^{infty}(x-b)e^{rac {-(x-b)^2}{2c^2}}dx+abint_{-infty}^{infty}e^{rac {-(x-b)^2}{2c^2}}dx

ГасГ.ca):-(..)dr + abl.부부.dr

=2ac^2int_{-infty}^{infty}left (rac {x-b}{csqrt {2}} ight )e^{-left (rac {x-b}{csqrt 2} ight )^2}dleft ( rac {x-b}{csqrt {2}} ight )+abint_{-infty}^{infty}e^{rac {-(x-b)^2}{2c^2}}dx

=2ac^2int_{-infty}^{infty}te^{-t^2}dt+acbsqrt 2int_{-infty}^{infty}e^{-left (rac {x-b}{csqrt 2} ight )^2}dleft (rac {x-b}{csqrt 2} ight ) (1st term is zero since the function is odd.)

=2acbsqrt 2int_{0}^{infty}e^{-left (rac {x-b}{csqrt 2} ight )^2}dleft (rac {x-b}{csqrt 2} ight )

=rac {1}{csqrt {2pi}}cbsqrt {2pi}

mathbf{<x>=b}

(iii)

<x^2>=int_{-infty}^{infty}x^2 ho(x)dx=aint_{-infty}^{infty}x^2e^{rac {-(x-b)^2}{2c^2}}dx

rac {x-b}{csqrt 2}=t

dx=csqrt 2 dt

<x^2>=acsqrt 2int_{-infty}^{infty}(csqrt 2t+b)^2e^{-t^2}dt

=acsqrt 2left (2c^2int_{-infty}^{infty}t^2e^{-t^2}dt +b^2int_{-infty}^{infty}e^{-t^2}dt +2sqrt 2bcint_{-infty}^{infty}te^{-t^2}dt ight )

=acsqrt 2left (2c^2int_{-infty}^{infty}t^2e^{-t^2}dt +b^2int_{-infty}^{infty}e^{-t^2}dt ight )

=acsqrt 2left (2c^2rac {sqrt {pi}}{2}+b^2sqrt {pi} ight )

=acleft (c^2+b^2 ight )sqrt {2pi}

=rac {1}{csqrt {2pi}}cleft (c^2+b^2 ight )sqrt {2pi}

<x^2>=c^2+b^2

(iv)

sd=sqrt {<x^2>-<x>^2}

sd=sqrt {c^2+b^2-b^2}

sd=sqrt {c^2}

mathbf{sd=c}

(v)

メ1 Α,

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