Finding a distribution function of a random variable
$begingroup$
Random variable $xi : (Omega, mathcal{F}, mathbb{P}) rightarrow mathbb{R}$ has Laplace distribution with density function $f_{xi} (x)= frac{1}{2}exp {-|x| }$ where $x in mathbb{R}$. I need to find a distribution function of random variable $xi^2$.
I have no idea what to do. But I think I could find a distribution function of $F_{xi}(x)$ because I have a density function and I know that $$F_{xi}(x)=int_{-infty}^xfrac{1}{2}exp {-|u| } du$$
So I found that
when $x<0$ we have $$F_{xi}(x)=frac{e^x}{2}$$
when $x geq 0$ we have $$F_{xi}(x)=1-frac{e^{-x}}{2}$$ but what I have to do next?
probability probability-theory probability-distributions random-variables
asked Dec 8 '18 at 13:25
AtstovasAtstovas
1089
$endgroup$
add a comment |
$begingroup$
Random variable $xi : (Omega, mathcal{F}, mathbb{P}) rightarrow mathbb{R}$ has Laplace distribution with density function $f_{xi} (x)= frac{1}{2}exp {-|x| }$ where $x in mathbb{R}$. I need to find a distribution function of random variable $xi^2$.
I have no idea what to do. But I think I could find a distribution function of $F_{xi}(x)$ because I have a density function and I know that $$F_{xi}(x)=int_{-infty}^xfrac{1}{2}exp {-|u| } du$$
So I found that
when $x<0$ we have $$F_{xi}(x)=frac{e^x}{2}$$
when $x geq 0$ we have $$F_{xi}(x)=1-frac{e^{-x}}{2}$$ but what I have to do next?
probability probability-theory probability-distributions random-variables
asked Dec 8 '18 at 13:25
AtstovasAtstovas
1089
$endgroup$
add a comment |
$begingroup$
Random variable $xi : (Omega, mathcal{F}, mathbb{P}) rightarrow mathbb{R}$ has Laplace distribution with density function $f_{xi} (x)= frac{1}{2}exp {-|x| }$ where $x in mathbb{R}$. I need to find a distribution function of random variable $xi^2$.
I have no idea what to do. But I think I could find a distribution function of $F_{xi}(x)$ because I have a density function and I know that $$F_{xi}(x)=int_{-infty}^xfrac{1}{2}exp {-|u| } du$$
So I found that
when $x<0$ we have $$F_{xi}(x)=frac{e^x}{2}$$
when $x geq 0$ we have $$F_{xi}(x)=1-frac{e^{-x}}{2}$$ but what I have to do next?
probability probability-theory probability-distributions random-variables
asked Dec 8 '18 at 13:25
AtstovasAtstovas
1089
$endgroup$
Random variable $xi : (Omega, mathcal{F}, mathbb{P}) rightarrow mathbb{R}$ has Laplace distribution with density function $f_{xi} (x)= frac{1}{2}exp {-|x| }$ where $x in mathbb{R}$. I need to find a distribution function of random variable $xi^2$.
I have no idea what to do. But I think I could find a distribution function of $F_{xi}(x)$ because I have a density function and I know that $$F_{xi}(x)=int_{-infty}^xfrac{1}{2}exp {-|u| } du$$
So I found that
when $x<0$ we have $$F_{xi}(x)=frac{e^x}{2}$$
when $x geq 0$ we have $$F_{xi}(x)=1-frac{e^{-x}}{2}$$ but what I have to do next?
probability probability-theory probability-distributions random-variables
probability probability-theory probability-distributions random-variables
asked Dec 8 '18 at 13:25
AtstovasAtstovas
1089
asked Dec 8 '18 at 13:25
AtstovasAtstovas
1089
edited Dec 8 '18 at 13:41
Atstovas
asked Dec 8 '18 at 13:25
AtstovasAtstovas
1089
asked Dec 8 '18 at 13:25
AtstovasAtstovas
1089
asked Dec 8 '18 at 13:25
AtstovasAtstovas
1089
1089
add a comment |
add a comment |
1 Answer
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$begingroup$
You can find an expression for $F_{xi^2}(x)$ now.
Evidently $F_{xi^2}(x)=0$ for $xleq 0$, and for $x>0$ we have:$$F_{xi^2}(x)=P(xi^2leq x)=P(-sqrt xleqxileqsqrt x)=F_{xi}(sqrt x)-F_{xi}(-sqrt x)$$
If next to the CDF you also want a PDF then find it as derivative of the CDF.
answered Dec 8 '18 at 14:23
drhabdrhab
100k544130
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1 Answer
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1 Answer
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$begingroup$
You can find an expression for $F_{xi^2}(x)$ now.
Evidently $F_{xi^2}(x)=0$ for $xleq 0$, and for $x>0$ we have:$$F_{xi^2}(x)=P(xi^2leq x)=P(-sqrt xleqxileqsqrt x)=F_{xi}(sqrt x)-F_{xi}(-sqrt x)$$
If next to the CDF you also want a PDF then find it as derivative of the CDF.
answered Dec 8 '18 at 14:23
drhabdrhab
100k544130
$endgroup$
add a comment |
$begingroup$
You can find an expression for $F_{xi^2}(x)$ now.
Evidently $F_{xi^2}(x)=0$ for $xleq 0$, and for $x>0$ we have:$$F_{xi^2}(x)=P(xi^2leq x)=P(-sqrt xleqxileqsqrt x)=F_{xi}(sqrt x)-F_{xi}(-sqrt x)$$
If next to the CDF you also want a PDF then find it as derivative of the CDF.
answered Dec 8 '18 at 14:23
drhabdrhab
100k544130
$endgroup$
add a comment |
$begingroup$
You can find an expression for $F_{xi^2}(x)$ now.
Evidently $F_{xi^2}(x)=0$ for $xleq 0$, and for $x>0$ we have:$$F_{xi^2}(x)=P(xi^2leq x)=P(-sqrt xleqxileqsqrt x)=F_{xi}(sqrt x)-F_{xi}(-sqrt x)$$
If next to the CDF you also want a PDF then find it as derivative of the CDF.
answered Dec 8 '18 at 14:23
drhabdrhab
100k544130
$endgroup$
You can find an expression for $F_{xi^2}(x)$ now.
Evidently $F_{xi^2}(x)=0$ for $xleq 0$, and for $x>0$ we have:$$F_{xi^2}(x)=P(xi^2leq x)=P(-sqrt xleqxileqsqrt x)=F_{xi}(sqrt x)-F_{xi}(-sqrt x)$$
If next to the CDF you also want a PDF then find it as derivative of the CDF.
answered Dec 8 '18 at 14:23
drhabdrhab
100k544130
answered Dec 8 '18 at 14:23
drhabdrhab
100k544130
answered Dec 8 '18 at 14:23
drhabdrhab
100k544130
answered Dec 8 '18 at 14:23
drhabdrhab
100k544130
100k544130
add a comment |
add a comment |
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