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$X$ is a continuous random variable with PDF $$f(x) = c\theta^{|x|} \quad \text{ for } -\infty<x<\infty,$$ where $0<\theta<1$ is a parameter and $c$ is a constant.

Derive and expression for $c$ in terms of $\theta$.

Well we have that $$I = \int^{\infty}_{-\infty} c\theta^{|x|} \,\rm dx = 1$$

Splitting this integral up:

\begin{align} I &= \int^{0}_{-\infty} c\theta^{-x} \, \rm dx + \int^{\infty}_{0} c\theta^{x} \, \rm dx - c \\ &= \left[-c\theta^{-x}\right]^{0}_{-\infty} + \left[c\theta^{x}\right]^{\infty}_{0} - c \\ &= [-c - \lim_{n\to-\infty} \theta^{-n}] + [\lim_{n\to\infty} \theta^{n} - c] - c \end{align}

Recall that $0<\theta<1$ hence both limits are zero so

$$I = -3c = 1 \iff c = -\frac{1}{3}$$

But this isn't really in terms of theta unless we include $\theta^{0}$ so is this correct?

user2850514
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2 Answers2

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The $-c$ in your first line is erroneous.

$f(x)$ makes sense as a distribution only for $0 < \theta < 1$ and then $$ 1 = 2c\int_0^\infty e^{x \ln \theta}dx = \left.\frac{2c}{\ln\theta}\theta^x\right|_0^\infty = -\frac{2c}{\ln\theta} $$ (note that the value of $\theta^x$ is zero at $x=\infty$ and one at $x=0$).

So your answer is $$ c = -\frac{\ln \theta}{2} $$

Mark Fischler
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  • That constant still doesn't make sense for the next part of the question either. This part is: Show $Pr(X\le x) = \theta^{-x}/2$ if $x<0$. Using your above constant we get $Pr(X\le x) = \theta^{-x}\ln{\theta}/2$. – user2850514 Oct 22 '14 at 21:17
  • "Using your above constant we get..." No, we get $\Pr(X≤x)=θ^{−x}/2$ if $x<0$. – Did Oct 22 '14 at 21:47
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You have that $$\begin{align*}1&=\int_{-\infty}^{+\infty}f(x)\, \rm dx=c\int^{0}_{-\infty} \theta^{-x} \, \rm dx+c\int^{\infty}_{0} \theta^{x} \, \rm dx=2c\int_{0}^{\infty}θ^x\, \rm dx =\\&=2c\left[\dfrac{θ^x}{\ln θ}\right]_{0}^{\infty}=2c\left[0-\dfrac{1}{\ln θ}\right]=-\dfrac{2}{\ln θ}c \end{align*}$$ which implies that $$c=-\dfrac{\ln θ}{2}$$ confirming the result in the othe answer. Hence, for $x<0$ the distribution function is $$F_X(x)=P(X\le x)=\int_{-\infty}^{x}f(t)dt=\int_{-\infty}^{x}-\dfrac{\ln θ}{2}θ^{-t}dt=-\dfrac{\ln θ}{2}\left[\dfrac{θ^{-t}}{\ln θ}\right]_{-\infty}^{x}=\dfrac{θ^{-x}}{2}$$

Jimmy R.
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