MHB Transformation of Random Variable

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The discussion centers on the validity of defining a uniform distribution for a random variable X in the range [0, Y], where Y follows a geometric distribution with mean alpha. Participants explore whether this definition holds true and seek the probability density function (pdf) for the transformation Y-X. The geometric distribution is defined with parameter p, leading to specific calculations for expected values and probabilities. The derived pdf of Y is linked to the cumulative distribution function of Y, which incorporates the uniform distribution of X. The conversation emphasizes the mathematical relationships between these distributions and transformations.
hemanth
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If X is a random variable distributed uniformly in [0, Y], where Y is geometric with mean alpha.
i) Is this definition valid for uniform distribution ?
ii) If it is valid, what is the pdf of the transformation Y-X?
 
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hemanth said:
If X is a random variable distributed uniformly in [0, Y], where Y is geometric with mean alpha.
i) Is this definition valid for uniform distribution ?
ii) If it is valid, what is the pdf of the transformation Y-X?

If Y is geomtrically distributed with parameter p, that means that... $$P \{ Y = n\} = p\ (1-p)^{n}\ (1)$$

... and...

$$E \{Y\} = \sum_{n=0}^{\infty} n\ p\ (1-p)^{n} = \frac{1-p}{p}\ (2)$$

If X is uniformely distributed in [0,Y], then is...

$$ P \{Y < x \} = p + p\ \sum_{n=1}^{\infty} \varphi_{n} (x)\ (1-p)^{n}\ (3)$$

... where...

$$\varphi_{n} (x)=\begin{cases}\ 0 & \text{if}\ x < 0 \\ \frac{x}{n} &\text{if}\ 0 \le x \le n \\ 1 &\text{if} x> n \end{cases}\ (4)
$$

Of course the p.d.f. of Y is the derivative of (3). I don't know if all that answers the question i) ... Kind regards $\chi$ $\sigma$
 
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