Applied Stochastic Processes: characteristic functions

[grandy]

Find characteristic functions of
1. The random variable X uniformly distributed on[-1..1]
2. The random variable Y distributed exponentially (with exponent λ)
3. The random variable Z=X+Y

 

[chisigma]

Find characteristic functions of
1. The random variable X uniformly distributed on[-1..1]
2. The random variable Y distributed exponentially (with exponent λ)
3. The random variable Z=X+Y

By definition the characteristic function of a r.v. X is...


$\displaystyle \varphi_{X} (t) = E\ \{e^{i\ t\ X}\} = \int_{ - \infty}^{+ \infty} f_{X} (x)\ e^{i\ t\ x}\ d x\ (1)$

For the point 1. is then...

$\displaystyle \varphi_{X} (t) = \frac{1}{2}\ \int_{-1}^{1} e^{i\ t\ x}\ dx = \frac{\sin t}{t}\ (2)$

The point 2. can be solved in similar way and the task is left to You as try. The point 3. requires the computation of $f_{X + Y} (x)$ and it will be done in next post...

Kind regards

$\chi$ $\sigma$

 

[grandy]

In (1) what is the value of Fx(x)?
From point(1) to (2), How did you go, is it by integrating?
In point 2 how did you get sint/t. Is this final answer for qs1.1 that I need to get?

 

[chisigma]

In (1) what is the value of Fx(x)?...

The (1) is a general formula that holds for any $f_{X}(x)$...


From point(1) to (2), how did you go, is it by integrating?...

Yes!...


In point 2 how did you get sint/t. Is this final answer for qs1.1 that I need to get?...

The function $\displaystyle \varphi (t) = \frac{\sin t}{t}$ is the result of integration... that is also the final answer!...

Kind regards

$\chi$ $\sigma$

 

[grandy]

Using 1/ b-a of the limit -1 to 1, I got fx (x) =1/2
But how to integrate e^(itx )in the limi -1 to 1 to get sin t/ t.
Any hint please

 

[grandy]

Point 1.i got= sint/t
point 2. I found out = λ/(it- λ)
would please help me to do point 3.

 

[chisigma]

Point 1.i got= sint/t
point 2. I found out = λ/(it- λ)
would please help me to do point 3.

Your solution of the point 2. is correct...

$\displaystyle \varphi_{X} (t) = \lambda\ \int_{0}^{\infty} e^{(i\ t - \lambda)\ x}\ d x = \frac{\lambda}{i\ t - \lambda}\ (1)$

The point 3. requires the evaluation of the p.d.f. of Z = X + Y. If X has p.d.f. $\displaystyle f_{X} (x)$ and Y has p.d.f. $\displaystyle f_{Y} (x)$ then Z = X + Y has p.d.f. ...

$\displaystyle f_{Z} (x) = f_{X} (x) * f_{Y} (x) = \int_{- \infty}^{+ \infty} f_{X} (\xi)\ f_{Y} (x - \xi)\ d \xi\ (2)$

The operation in (2) is called convolution and it is efficiently performed using the Laplace Transform. We have...

$\displaystyle \mathcal{L} \{ f_{X} (x)\} = \frac{\sinh s}{s}\ (3)$

$\displaystyle \mathcal{L} \{ f_{Y} (x)\} = \frac{\lambda}{s + \lambda}\ (4)$

... so that is...


$\displaystyle \mathcal {L} \{f_{Z} (x)\} = \frac{\lambda\ \sinh s}{s\ (s + \lambda)} \implies f_{Z} (x) = \frac{1 - e^{- \lambda\ x}}{2}\ \{\mathcal {U} (x + 1) - \mathcal{U} (x-1)\}\ (5)$

Are You able to proceed alone?...

Kind regards

$\chi$ $\sigma$

 

[grandy]

NO, I cannot proceed alone. im confused.
how did you get point 3, 4 and 5. is point 5 the final answer?
and how did you change to to s in point 3 and 4

 

[chisigma]

NO, I cannot proceed alone. im confused.
how did you get point 3, 4 and 5. is point 5 the final answer?
and how did you change to to s in point 3 and 4

As preliminary consideration I have to precise that if someone wants to operate in advanced probability, knowledge of advanced calculus, that includes complex variable function theory, convolution, Laplace and Fourier Transforms is essential. The 'final answer' to point 3 is the evaluation of the charactristic function of the r.v. Z, the p.d.f. of which is given by...

$\displaystyle f_{Z} (x) = \frac{1 - e^{- \lambda\ x}}{2}\ \{\mathcal{U} (x+1) - \mathcal{U} (x-1) \}\ (1)$

By definition is...

$\displaystyle \varphi_{Z} (t) = E \{e^{i\ t\ Z}\} = \int_{-1}^{1} \frac{1- e^{- \lambda\ x}}{2}\ e^{i\ t\ x}\ dx = \frac{\sin t}{t} - \frac{\sinh (\lambda - i\ t)}{\lambda - i\ t}\ (2)$

Kind regards

$\chi$ $\sigma$