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$X\sim U(1,3)$. Verify that X has cdf $F_X(x) = 2(x − 1)$ for $x \epsilon(1, 3)$ and thus that $F^{−1}_X (y) = 2y +1$ for $y \epsilon (0, 1)$.

My attempt for $F_X(x)=\int_{-\infty}^{\infty}\frac{1}{b-a}dx=\int_{1}^{x}\frac{1}{3-1}dx=\frac{1}{2}\int_{1}^{x}dx=\frac{1}{2}[x-1]$

But the result is $F_X(x) = 2(x − 1)$.

I couldn't solve $F^{−1}_X (y)$

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2 Answers 2

up vote 1 down vote accepted

Surely the question that says $F_{X}(x) = 2(x-1)$ is wrong, since if you substitute 3 for $x$ you are getting $4$ as the answer.

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Thank you. My first part has been solved. –  harry Sep 22 '13 at 6:43
    
Second part is straight forward from your first answer. Invert the expression to get the value of $x$. To be more clear, the 2nd part asks for the value of $x$ when $y$ is the given CDF value. So $F_{X}(x)=\frac{x-1}{2}$ and the value of $x$ which gives $y$ as the CDF value is simply $2y+1.$ –  Sudarsan Sep 22 '13 at 6:45

I believe that is a typo because your answer is indeed correct. For the second part you can set $x = (y-1)\frac{1}{2}$ and just solve for y. This will give you your inverse.

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Yes, there is a theorem that says that if U is uniformly distributed over the interval (0,1), then $X = F_X^{-1}(U)$ has c.d.f $F_X(x)$. So the inverse is distributed over (0,1). To see this $P(X\leq x) = P(F_X^{-1}(U)\leq x) = P(U\leq F_X(x)) = F_X(x)$. –  RDizzl3 Sep 22 '13 at 6:57
    
Thank you. Your above comment is due to my another question in the comment[pardon me, which i have erased before posting your comment.]. Now The second part is clearer. –  harry Sep 22 '13 at 7:07

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