uniform distribution on [0,1] find function
$begingroup$
Consider $X∼unif [0,1]$. Find a function $g: mathbb{R} longrightarrow mathbb{R}$, such that g(X) has pdf $f(t) = begin{cases} {t+1}, & text{$-1 leq tleq 0$} \ {1-t}, &
text{$0<tleq 1$}end{cases}$.
Can you help me, please? I do not know what I have to do.
probability probability-theory random-variables uniform-distribution density-function
asked Jan 3 at 17:16
tommy_mtommy_m
1915
$endgroup$
add a comment |
$begingroup$
Consider $X∼unif [0,1]$. Find a function $g: mathbb{R} longrightarrow mathbb{R}$, such that g(X) has pdf $f(t) = begin{cases} {t+1}, & text{$-1 leq tleq 0$} \ {1-t}, &
text{$0<tleq 1$}end{cases}$.
Can you help me, please? I do not know what I have to do.
probability probability-theory random-variables uniform-distribution density-function
asked Jan 3 at 17:16
tommy_mtommy_m
1915
$endgroup$
add a comment |
$begingroup$
Consider $X∼unif [0,1]$. Find a function $g: mathbb{R} longrightarrow mathbb{R}$, such that g(X) has pdf $f(t) = begin{cases} {t+1}, & text{$-1 leq tleq 0$} \ {1-t}, &
text{$0<tleq 1$}end{cases}$.
Can you help me, please? I do not know what I have to do.
probability probability-theory random-variables uniform-distribution density-function
asked Jan 3 at 17:16
tommy_mtommy_m
1915
$endgroup$
Consider $X∼unif [0,1]$. Find a function $g: mathbb{R} longrightarrow mathbb{R}$, such that g(X) has pdf $f(t) = begin{cases} {t+1}, & text{$-1 leq tleq 0$} \ {1-t}, &
text{$0<tleq 1$}end{cases}$.
Can you help me, please? I do not know what I have to do.
probability probability-theory random-variables uniform-distribution density-function
probability probability-theory random-variables uniform-distribution density-function
asked Jan 3 at 17:16
tommy_mtommy_m
1915
asked Jan 3 at 17:16
tommy_mtommy_m
1915
asked Jan 3 at 17:16
tommy_mtommy_m
1915
asked Jan 3 at 17:16
tommy_mtommy_m
1915
asked Jan 3 at 17:16
tommy_mtommy_m
1915
1915
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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Hint/Guide
Compute the cdf $F$ corresponding to $f$ (i.e. $F(x)=int_{-infty}^x f(t), dt)$ and use the fact that $F^{-1}(X)$ has the same distribution as $F$. So we can take $g=F^{-1}$.
answered Jan 3 at 17:35
Foobaz JohnFoobaz John
22.6k41452
$endgroup$
$begingroup$
Thanks for your answer. So I have $ F(x) = begin{cases} 0 & x <-1\ frac12 + x + frac{x^2}{2} & x in[-1,0]\ frac12 + x - frac{x^2}{2} & x in(0,1]\ 0 & x > 1end{cases}$. Is that correct? Now I have to calculate $F^{-1}(x)$ for every case, right?
$endgroup$
– tommy_m
Jan 3 at 18:36
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Hint/Guide
Compute the cdf $F$ corresponding to $f$ (i.e. $F(x)=int_{-infty}^x f(t), dt)$ and use the fact that $F^{-1}(X)$ has the same distribution as $F$. So we can take $g=F^{-1}$.
answered Jan 3 at 17:35
Foobaz JohnFoobaz John
22.6k41452
$endgroup$
$begingroup$
Thanks for your answer. So I have $ F(x) = begin{cases} 0 & x <-1\ frac12 + x + frac{x^2}{2} & x in[-1,0]\ frac12 + x - frac{x^2}{2} & x in(0,1]\ 0 & x > 1end{cases}$. Is that correct? Now I have to calculate $F^{-1}(x)$ for every case, right?
$endgroup$
– tommy_m
Jan 3 at 18:36
add a comment |
$begingroup$
Hint/Guide
Compute the cdf $F$ corresponding to $f$ (i.e. $F(x)=int_{-infty}^x f(t), dt)$ and use the fact that $F^{-1}(X)$ has the same distribution as $F$. So we can take $g=F^{-1}$.
answered Jan 3 at 17:35
Foobaz JohnFoobaz John
22.6k41452
$endgroup$
$begingroup$
Thanks for your answer. So I have $ F(x) = begin{cases} 0 & x <-1\ frac12 + x + frac{x^2}{2} & x in[-1,0]\ frac12 + x - frac{x^2}{2} & x in(0,1]\ 0 & x > 1end{cases}$. Is that correct? Now I have to calculate $F^{-1}(x)$ for every case, right?
$endgroup$
– tommy_m
Jan 3 at 18:36
add a comment |
$begingroup$
Hint/Guide
Compute the cdf $F$ corresponding to $f$ (i.e. $F(x)=int_{-infty}^x f(t), dt)$ and use the fact that $F^{-1}(X)$ has the same distribution as $F$. So we can take $g=F^{-1}$.
answered Jan 3 at 17:35
Foobaz JohnFoobaz John
22.6k41452
$endgroup$
Hint/Guide
Compute the cdf $F$ corresponding to $f$ (i.e. $F(x)=int_{-infty}^x f(t), dt)$ and use the fact that $F^{-1}(X)$ has the same distribution as $F$. So we can take $g=F^{-1}$.
answered Jan 3 at 17:35
Foobaz JohnFoobaz John
22.6k41452
answered Jan 3 at 17:35
Foobaz JohnFoobaz John
22.6k41452
answered Jan 3 at 17:35
Foobaz JohnFoobaz John
22.6k41452
answered Jan 3 at 17:35
Foobaz JohnFoobaz John
22.6k41452
22.6k41452
$begingroup$
Thanks for your answer. So I have $ F(x) = begin{cases} 0 & x <-1\ frac12 + x + frac{x^2}{2} & x in[-1,0]\ frac12 + x - frac{x^2}{2} & x in(0,1]\ 0 & x > 1end{cases}$. Is that correct? Now I have to calculate $F^{-1}(x)$ for every case, right?
$endgroup$
– tommy_m
Jan 3 at 18:36
add a comment |
$begingroup$
Thanks for your answer. So I have $ F(x) = begin{cases} 0 & x <-1\ frac12 + x + frac{x^2}{2} & x in[-1,0]\ frac12 + x - frac{x^2}{2} & x in(0,1]\ 0 & x > 1end{cases}$. Is that correct? Now I have to calculate $F^{-1}(x)$ for every case, right?
$endgroup$
– tommy_m
Jan 3 at 18:36
$begingroup$
Thanks for your answer. So I have $ F(x) = begin{cases} 0 & x <-1\ frac12 + x + frac{x^2}{2} & x in[-1,0]\ frac12 + x - frac{x^2}{2} & x in(0,1]\ 0 & x > 1end{cases}$. Is that correct? Now I have to calculate $F^{-1}(x)$ for every case, right?
$endgroup$
– tommy_m
Jan 3 at 18:36
$begingroup$
Thanks for your answer. So I have $ F(x) = begin{cases} 0 & x <-1\ frac12 + x + frac{x^2}{2} & x in[-1,0]\ frac12 + x - frac{x^2}{2} & x in(0,1]\ 0 & x > 1end{cases}$. Is that correct? Now I have to calculate $F^{-1}(x)$ for every case, right?
$endgroup$
– tommy_m
Jan 3 at 18:36
add a comment |
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