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- Thread starter grandy
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- #1

- Feb 13, 2012

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

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

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

$\chi$ $\sigma$

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- Feb 13, 2012

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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

$\chi$ $\sigma$

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- Feb 13, 2012

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Your solution of the point 2. is correct...Point 1.i got= sint/t

point 2. I found out = λ/(it- λ)

would please help me to do point 3.

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

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

$\chi$ $\sigma$

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- Feb 13, 2012

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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...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

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

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

$\chi$ $\sigma$