Question

Let X be an exponential random variable with parameter 2. Find the CDF of X . Express your answer in terms of x using standard notation. Use 'e' for the base of the natural logarithm (e.g., enter e^(-3*x) for e−3x ). a) For x≤0 , FX(x)= unanswered b) For x>0 , FX(x)=

Answer 1

Answer:

$F(x,\lambda) = 1-e^{2x}$     for      $x\geq0$

and   $F(x,\lambda) = 0$    for   $x <0$

Step-by-step explanation:

Remember that the CDF of an exponental random variable is given by

$F(x,\lambda) = \left \{ {{1-e^{\lambda x} \ \ \ \ \ x\geq 0} \atop {0} \ \ \ \ \ \ \ \ \ \ \ x<0} \right.$

where   $\lambda$   is the parameter.  In this case    $\lambda = 2$  so, for this case

$F(x,\lambda) = \left \{ {{1-e^{2x} \ \ \ \ \ x\geq 0} \atop {0} \ \ \ \ \ \ \ \ \ \ \ x<0} \right.$

Answer 2

Part(a) For $x\leq 0,F_{x}\left ( x \right )=0$

Part(b) For $x > 0,F_{x}\left ( x \right )=1-e^{-2x}$

Probability density function:

The probability density function (PDF) is used to define the random variable’s probability coming within a distinct range of values, as opposed to taking on anyone's value.

Part(a):

Given that,

$x\sim Exp\left ( \lambda =2 \right )$

$\therefore$The PDF of $x$ is

$\Rightarrow f_{x}\left ( x \right )=\left\{\begin{matrix}\lambda e^{-\lambda x}=2.e^{-2x} \ ;for,x > 0 & \\ 0 \ ;otherwise & \end{matrix}\right.$

Define, $F_{x}\left ( x \right )=p\left ( x\leq x \right )=cdf \ of \ x$

$\Rightarrow$So, when $x\leq 0;$we have,

$F_{x}\left ( x \right )=p\left ( x\leq x \right )=0$

Part(b):

when $x > 0;$ we have,

$x\leq 0,F_{x}\left ( x \right )=0 \\ x > 0,F_{x}\left ( x \right )=1-e^{-2x}x\sim Exp\left ( \lambda =2 \right ) \\ \Rightarrow f_{x}\left ( x \right )=\left\{\begin{matrix}\lambda e^{-\lambda x}=2.e^{-2x} \ ;for,x > 0 & \\ 0 \ ;otherwise & \end{matrix}\right. \\ F_{x}\left ( x \right )=p\left ( x\leq x \right )=cdf \ of \ x \\ F_{x}\left ( x \right )=p\left ( x\leq x \right )=0 \\ \x > 0;$

$F_{x}\left ( x \right )=p\left ( x\leq x \right ) \\ =\int_{0}^{x}d_{x}\left ( t \right )dt \\ =\int_{0}^{x}2e^{-2t}dt\\=(1-e^{-2x})$

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