Moments and Moment Generating Function of Standard Laplace Distribution

Moments of Standard Laplace Distribution:

Let X ~ L (0,1), ie. a random variable X follows a standard Laplace distribution, then the r^th moment about origin of X is given by

$$\mu_r^{'} \ = \ E (X^r)$$

$$= \ \int_{-\infty}^{\infty} \ f(x) \ dx$$

$$= \ \frac{1}{2} \ \int_{-\infty}^{\infty} \ x^r \ {e^-}^{|x|} \ dx$$

$$= \ \frac{1}{2} \ \left [ \ \int_{-\infty}^{0} \ x^r \ e^x \ dx \ + \ \int_{0}^{\infty} \ x^r \ {e^-}^{x} \ dx \ \right ]$$

Here,

$$\int_{-\infty}^{0} \ x^r \ e^x \ dx \ = \ \int_{-\infty}^{0} \ (-y)^r \ {e^-}^{y} \ (-dy) \ \ [ \ putting \ x \ = \ - \ y, \ so \ that \ dx \ = \ - \ dy \ ]$$

$$= \ (-1)^r \ \int_{0}^{\infty} \ y^r \ {e^-}^{y} \ dy$$

$$= \ (-1)^r \ \Gamma(r+1)$$

$$= \ (-1)^r \ r!$$

$$\therefore \ \mu_r^{'} \ = \ \frac{1}{2} \ [ (-1)^r \ r! \ + \ r! \ ]$$

$$\mu_r^{'} \ = \ \frac{(-1)^r + 1}{2} \ r!$$

Thus, we get first four raw moments about origin as

$$\mu_1^{'} \ = \ 0$$

$$\mu_2^{'} \ = \ 2$$

$$\mu_3^{'} \ = \ 0$$

$$\mu_4^{'} \ = \ 24$$

Since μ₁^'= 0, the cental moments of the standard Laplace distribution coincide with the raw moments. Then,

$$\mu_1 \ = \ 0$$

$$\mu_2 \ = \ \mu_2^{'} \ = \ 2,$$

$$\mu_3 \ = \ \mu_3^{'} \ = \ 0,$$

$$\mu_4 \ = \ \mu_4^{'} \ = \ 24$$

Hence, the standard Laplace distribution has mean = 0 and variance = 2.

Also, the moment coefficient of skewness is

$$\beta_1 \ = \ \frac{\mu_3^{2}}{\mu_2^{3}} \ = 0$$

$$\Rightarrow \ \gamma_1 \ = \ \sqrt \beta_1 \ = \ 0$$

and the moment coefficient of kurtosis is

$$\beta_2 \ = \ \frac{\mu_4}{\mu_2^{2}} \ = \ \frac{24}{2^2} \ = \ 6$$

$$\Rightarrow \ \gamma_2 \ = \ \beta_2 \ - \ 3 \ = \ 6 \ - \ 3 \ = \ 3$$

Hence, it can be concluded from the values ofγ₁ andγ₂ that the standard Laplace distribution is symmetrical but it is leptokurtic.

Moment Generating Function of Standard Laplace Distribution:

The moment generating function of the double exponential distribution is given by

$$M_X(t) \ = \ E({e^t}^{x}) \ = \ \int_{-\infty}^{\infty} \ {e^t}^{x} \ \frac{1}{2} \ {e^-}^{|x|} \ dx$$

$$= \ \frac{1}{2} \ \left [ \ \int_{-\infty}^{0} \ {e^t}^{x} \ e^x \ dx \ + \ \int_0^{\infty} \ {e^t}^{x} \ {e^-}^{x} \ dx \ \right ]$$

$$= \ \frac{1}{2} \ \left [ \ \frac{{e^(}^{1+t)x}}{1+t} |_{-\infty}^{0} \ - \ \frac{{e^-}^{(1-t)x}}{1-t} |_0^{\infty} \ \right ]$$

$$= \ \frac{1}{2} \ \left [ \ \frac{1}{1+t} \ + \ \frac{1}{1-t} \ \right ]$$

$$\therefore \ \ M_x(t) \ = \ \frac{1}{1-t^2}$$

Expanding M_x(t), we get

$$M_x(t) \ = \ {(1-t^2)^-}^{1}$$

$$= \ 1 \ + \ t^2 \ + \ t^4 \ + \ . \ . \ .$$
$$= \ 1 \ + \ 2 \frac{t^2}{2!} \ + \ 4! \frac{t^4}{4!} + \ . \ . \ .$$

$$\therefore \ \mu_r^{'} \ = \ coefficient \ of \ \frac{t^r}{r!} \ in \ M_x(t)$$

Hence, the first four raw moments of the Standard Laplace distribution are:

$$\mu_1^{'} \ = \ 0$$

$$\mu_2^{'} \ = \ 2$$

$$\mu_3^{'} \ = \ 0$$

$$\mu_4^{'} \ = \ 24$$

Characteristic Function of Standard Laplace Distribution

The characteristic function of standard Laplace Distribution is given by

$$\phi x (t) \ = \ E({e^i}^{tx}) \ = \ \int_{-\infty}^{\infty} \ {e^i}^{tx} \ f(x) \ dx$$

$$= \ \frac{1}{2} \ \int_{\infty}^{\infty} \ {e^i}^{tx} \ {e^-}^{|x|} \ dx$$

$$= \ \frac{1}{2} \ \left [ \ \int_{-\infty}^{0} \ {e^i}^{tx} \ e^x \ dx \ + \ \int_0^{\infty} \ {e^i}^{tx} \ {e^-}^{x} \ dx \ \right ]$$

$$\therefore \ \phi x (t) \ = \ \frac{1}{1 + t^2}$$

In order to find the moments, we expand the characteristic function, as

$$\phi x (t) \ = \ \frac{1}{1+t^2} \ = \ {(1+t^2)^-}^{1}$$

$$= \ 1 \ - \ t^2 \ + \ t^4 \ - \ t^6 \ + \ . \ . \ .$$
$$= \ 1 \ + \ (it)^2 \ + \ (it)^4 \ + \ (it)^6 \ + \ . \ . \ .$$

Since the characteristic function has only the terms of even powers of t, all the moments of odd order vavish. So,

$$\mu_1^{'} \ = \ 0$$

$$\mu_2^{'} \ = \ 2$$

$$\mu_3^{'} \ = \ 0$$

$$\mu_4^{'} \ = \ 24$$

Hence, mean = 0 and variance = 2.

Mean Deviation about mean of Laplace Distribution

The mean deviation about mean is given by

$$M.D. \ (0) \ = \ \int_{-\infty}^{\infty} \ |x-0| \ f(x) \ dx$$

$$= \ \frac{1}{2} \ \int_{-\infty}^{\infty} \ |x| \ {e^-}^{|x|} \ dx$$

$$= \ \frac{1}{2} \ \left [ \ \int_{\infty}^{0} \ (-x) \ e^x \ dx \ + \ \int_0^{\infty} x \ {e^-}^{x} \ dx \ \right ] $$

$$= \ \frac{1}{2} \ [ \ 0 + \ 1 \ + \ 1 \ ]$$

$$= \ 1$$

Interquartile Range of Standard Laplace Distribution

The first quartile Q₁ is given by

$$\int_{-\infty}^{Q_1} \ f(x) \ dx \ = \ \frac{1}{4}$$

$$\Rightarrow \ \int_{-\infty}^{Q_1} \ \frac{1}{2} \ {e^-}^{|x|} \ dx \ = \ \frac{1}{4}$$

$$\Rightarrow \ \int_{-\infty}^{Q_1} \ e^x \ dx \ = \ \frac{1}{4}$$

$$\Rightarrow \ frac{1}{2} \ \left [ \ e^x \ \right ]_{-\infty}^{Q_1} \ = \ \frac{1}{4}$$

$$\Rightarrow \ \frac{1}{2} \ {e^Q}^{_1} \ = \ \frac{1}{4}$$

$$\therefore \ Q_1 \ = \ - \ log_e \ 2$$

The third Quartile Q₃ is given by

$$\int_{-\infty}^{Q_3} \ f(x) \ dx \ = \ \frac{3}{4}$$

$$\Rightarrow \ \int_{-\infty}^{Q_3} \ \frac{1}{2} \ {e^-}^{|x|} \ dx \ = \ \frac{3}{4}$$

$$\Rightarrow \ 1 - \ {e^-}^{Q_3} \ + \ 1 \ = \ frac{3}{2}$$

$$\Rightarrow \ {e^-}^{Q_3} \ = \ \frac{1}{2}$$

$$\therefore \ Q_3 \ = \ log_e \ 2$$

The inter-quartile range of standard Laplace distribution is given by

Q₃ - Q₂ = log_e 2 - (-log_e 2)

= 2 log_e 2

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.