Characteristic Function Of Cauchy Distribution
Then the characteristic function of Cauchy distribution X is given by Φx(t) = E [eitX] = E [eit(μ+λz)] = eitμ E (eitλz) = eitμΦz(tλ) =eitμ .e-|tλ| ∴Φx(t) =eitμ-λ|t| ;λ > 0
Summary
Then the characteristic function of Cauchy distribution X is given by Φx(t) = E [eitX] = E [eit(μ+λz)] = eitμ E (eitλz) = eitμΦz(tλ) =eitμ .e-|tλ| ∴Φx(t) =eitμ-λ|t| ;λ > 0
Things to Remember
-
Then the characteristic function of Cauchy distribution X is given by
Φx(t) = E [eitX]
= E [eit(μ+λz)]
= eitμ E (eitλz)
= eitμΦz(tλ)
=eitμ .e-|tλ|
∴Φx(t) =eitμ-λ|t| ;λ > 0
- Cauchy distribution follows additive property
MCQs
No MCQs found.
Subjective Questions
No subjective questions found.
Videos
No videos found.

Characteristic Function Of Cauchy Distribution
Let X follows a Cauchy distribution with a parameterμ andλ i.e. X ~ (μ,λ ). To find the characteristic function of X, we first find the characteristic function of the stsndard Cauchy variate
$$Z \ = \ \frac{X \ - \ \mu}{\lambda}$$
The probability function of Z is given by
$$f(x) \ = \ \frac{1}{\pi} \ . \ \frac{1}{1+z^2} \ \ \ \ ; \ -\infty \ < \ z \ < \ \infty$$
The characteristic function of Z is given by
$$\phi_{z}(t) \ = \ E({e^i}^{tz})$$
$$= \ \int_{-\infty}^{\infty}\ {e^i}^{tz} \ f(z) \ dz$$
$$= \ \frac{1}{\pi} \ \int_{-\infty}^{\infty}\ \frac{1}{1+z^2} \ dz$$
To find Φz(t), we consider stsndard Laplace distribution given by the probability density function
$$f_1(z) \ = \ \frac{1}{2} \ {e^-}^{|z|}, \ \ \ \; \ -\infty \ < \ z \ < \ \infty$$
Then, the characteristic function of the standard Laplace variable is given by
$$\phi_1 (t) \ = \ \frac{1}{2} \ \int_{-\infty}^{\infty} \ {e^i}^{tz} \ {e^-}^{|z|} \ dz$$
$$= \\frac{1}{2} \ \int_{-\infty}^{\infty} \ (Coz \ tz \ + \ i \ sin \ tz) \ {e^-}^{|z|} \ dz$$\
$$ = \ \frac{1}{2} \ \int_{-\infty}^{\infty} \ Coz \ tz \ . \ {e^-}^{|z|} \ dz \ \ \ \ \ \ \ \ \ [\therefore \ The \ second \ integral \ being \ an \ odd \ function \ vanishes]$$
$$= \ \int_{o}^{\infty} \ Cos \ tz \ . \ {e^-}^{|z|} \ dz$$
Integrating by parts, we get,
$$ = \ \left [ \ -Cos \ tz \ . \ {e^-}^{z} \ \right ]_{0}^{\infty} \ -t \int_{o}^{\infty} \ sin \ tz \ . \ {e^-}^{z} \ dz$$
$$= \ 1 \ - \ t \ \int_{o}^{\infty} \ sin \ tz \ . \ {e^-}^{z} \ dz$$
$$ \ = \ 1 \ - \ t \ \left [ -{e^-}^{z} \ sin \ tz \ \mid_{0}^{\infty} \ + \ t \ \int_{0}^{\infty} \ coz \ tz \ {e^-}^{z} \ dz \ \right ]$$
$$= \ 1 \ - \ t^2 \ \int_{0}^{\infty} \ coz \ tz \ {e^-}^{z} \ dz$$
$$\therefore \int_{0}^{\infty} \ coz \ tz \ {e^-}^{z} \ dz = \ 1 \ - \ t^2 \ \int_{0}^{\infty} \ coz \ tz \ {e^-}^{z} \ dz$$
$$\Rightarrow \ \int_{0}^{\infty} \ coz \ tz \ {e^-}^{z} \ dz \ = \ \frac{1}{1 + t^2}$$
$$Hence, \phi_1(t) \ =\ \frac{1}{1 + t^2}.$$
This characteristic functionΦ1(t) is absolutely integral in (-∞,∞). Thus, by inversion theorem, the corresponding density function is
$$f_1 \ (z) \ = \ \frac{1}{2 \pi} \ \int_{-\infty}^{\infty} \ {e^-}^{itz} \ \phi_1 (t) \ dt$$
$$ or, \ \ \ \frac{1}{2} \ {e^-}^{|z|} \ = \ \frac{1}{2 \pi} \ \int_{-\infty}^{\infty} \ {e^-}^{itz} \ \frac{1}{1+t^2} \ dt$$
$$ or, \ \ \ {e^-}^{|z|} \ = \ \frac{1}{\pi} \ \int_{-\infty}^{\infty} \ \frac{{e^-}^{itz}}{1+t^2} \ dt \ \ \ \left [ \ \therefore \ Changing \ t \ to \ -t \ which \ does \not \ affect \ the \ value \ of \ the \ integral \ \right ]$$
On changing t and z, we get
$${e^-}^{|t|} \ = \ \frac{1}{\pi} \ \int_{-\infty}^{\infty} \ \frac{{e^-}^{itz}}{1+t^2} \ dz$$
$$ = \ \phi_z (t)$$
Hence, the characteristic function of standard Cauchy variate Z is
$$ \phi_z (t) \ = \ {e^-}^{|t|}$$
Then the characteristic function of Cauchy distribution X is given by
Φx(t) = E [eitX]
= E [eit(μ+λz)]
= eitμ E (eitλz)
= eitμΦz(tλ)
=eitμ .e-|tλ|
∴Φx(t) =eitμ-λ|t| ;λ > 0
Further, since |t| is not differentiable at t = 0, theΦx(t) is not differentiable at t = 0. Since rth moment of X is given by
$$\mu_{r}^{'} \ = \ \frac{1}{i^r} \ \left [ \frac{d^r \ \phi_x(t)}{dt^r} \ \right ]_{t=0}\ = \ \frac{\phi_x(0)}{i^r}$$
the moments of any ordher do not exist. Hence the mean and variance of Cauchy distribution do not exist. This property is a very peculiar one which is not found in other distributions discussed so far.
Additive Property of Cauchy Distribution
If X1 and X2 are two independent Cauchy varaites with parameters (μ1,λ1) and (μ2,λ2) respectively, then X1 + X2 is a Cauchy variate with parameters (μ1 +μ2,λ1 +λ2)
proof:
We have the characteristic function of X1 and X2 are given by
Φx1 (t) = eitμ1 -λ1 |t|
and Φx2(t) = eitμ2 -λ2 |t|
Since X1 and X2 are independent, the characteristic function of X1 + X2 is given by
Φx1 + x2 (t) =Φx1 (t) .Φx2(t)
∴Φx1 + x2 (t) = eit(μ1 +μ2) - (λ1 +λ2) |t|
which is the characteristic function of a Cauchy function of a Cauchy random variable with parameter(μ1 +μ2,λ1 +λ2). Hence by uniqueness theorem of characteristic functionX1 + X2~(μ1 +μ2,λ1 +λ2). Thus the Cauchy distribution follows additive property.
Bibliography
Sukubhattu N.P. (2013). Probability & Inference - II. Asmita Books Publishers & Distributors (P) Ltd., Kathmandu.
Larson H.J. Introduction to Probability Theory and Statistical Inference. WileyInternational, New York.
Lesson
continuious probablity distributions
Subject
Statistics
Grade
Bachelor of Science
Recent Notes
No recent notes.
Related Notes
No related notes.