TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW02.2

Continuous random variable

Let $X$ be a continuous random variable with the probability density function (pdf)

$f_{X}(x)={\begin{cases}\alpha e^{-x},\ x\in (0,\ln 2),\\0,\ {\text{otherwise}}\end{cases}}$

where $\alpha$ is a positive real constant and $\ln$ denotes the natural logarithm.

(a) Determine the value of $\alpha$.

(b) Find the cummulative distribution function (cdf) of $X$.

(c) Compute the first two moments of $X$.

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Lösungsvorschlag von Lessi

--Lessi 2024-02-07T13:04:11Z

a

Since $f_{X}$ is a continous density function we know that

${\begin{aligned}1&=\int _{-\infty }^{\infty }f_{X}(x)dx=\int _{0}^{\ln 2}\alpha e^{-x}dx\\1&=\alpha *(-e^{-\ln 2}-e^{0})=\alpha *(-{\frac {1}{2}}+1)\\\alpha &=2\end{aligned}}$

Now the pdf can be defined as: $f_{X}(x)={\begin{cases}2e^{-x},&x\in (0,\ln 2),\\0,&{\text{otherwise}}\end{cases}}$

b

We are only interested in the values for $0\leq x<\ln 2$ which is the Integral $\int _{0}^{x}2e^{-t}dt=-2e^{-x}+2$. So the cdf is defined as:

$F(X)={\begin{cases}0,&x<0,\\-2e^{-x}+2,&0\leq x<\ln 2,\\1,&X\geq \ln 2\end{cases}}$

c

The first moment is: $\mathbb {E} (X)=\int _{0}^{\ln 2}x2e^{-x}dx$ which is computed using partial integration with $f'=e^{-x},\ g=x$: $\mathbb {E} (X)=2*(-{\frac {1}{2}}\ln 2+\int _{0}^{\ln 2}e^{-x}dx)=2*(-{\frac {1}{2}}\ln 2-{\frac {1}{2}}+1)=1-\ln 2$

The second moment $\mathbb {E} (X^{2})=\int _{0}^{\ln 2}x^{2}2e^{-x}dx$ can be done similarly:

${\begin{aligned}\int _{0}^{\ln 2}x^{2}e^{-x}dx&=-{\frac {1}{2}}\ln ^{2}{2}+2\int _{0}^{\ln 2}e^{-x}xdx=-{\frac {\ln ^{2}2}{2}}+1-\ln 2\\\mathbb {E} (X^{2})&=2*(1-\ln 2-{\frac {\ln ^{2}2}{2}})\end{aligned}}$