uniform distribution on [0,1] find function












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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







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    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







    share|cite|improve this question

























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      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







      share|cite|improve this question













      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






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      asked 2 days ago









      tommy_m

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      1115






















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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 $g=F^{-1}$.






          share|cite|improve this answer





















          • 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?
            – tommy_m
            2 days ago











          Your Answer





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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 $g=F^{-1}$.






          share|cite|improve this answer





















          • 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?
            – tommy_m
            2 days ago
















          0














          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 $g=F^{-1}$.






          share|cite|improve this answer





















          • 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?
            – tommy_m
            2 days ago














          0












          0








          0






          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 $g=F^{-1}$.






          share|cite|improve this answer












          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 $g=F^{-1}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Foobaz John

          21.4k41351




          21.4k41351












          • 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?
            – tommy_m
            2 days ago


















          • 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?
            – tommy_m
            2 days ago
















          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?
          – tommy_m
          2 days ago




          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?
          – tommy_m
          2 days ago


















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