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Transform random numbers that are uniform on interval [0,1] into random numbers that are uniform on the interval [-11,17] I really don't know how to start I use lets F(x) = X <= x = x

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Let random variable $X$ be uniformly distributed on the interval $[0,1]$. We will find constants $a$ and $b$ such that the random variable $aX+b$ is uniformly distributed on $[-11,17]$.

The intuition is that if we find $a$ and $b$ such that $(a)(0)+b=-11$, and $(a)(1)+b=17$, that should do the job. Solving these two equations for $a$ and $b$, we find that $b=-11$ and $a=28$.

Intuition is perhaps not enough. Let us show that the random variable $Y$, where $Y=28X-11$, is indeed uniformly distributed in the interval $[-11,17]$. Note that $Y$ is just a scaling of $X$ followed by a shift. That should preserve uniform distribution.

So we want to prove that for any $y$ between $-11$ and $17$, we have $\Pr(Y\le y)=\frac{y-(-11)}{28}$, for this is what uniform distribution on $[-11,17]$ means.

For such a $y$, we have $$\Pr(Y\le y)=\Pr(28X-11\le y)=\Pr(28X\le y-(-11))=\Pr\left(X\le \frac{y-(-11)}{28}\right).$$

It is not hard to check that $\frac{y-(-11)}{28}$ lies between $0$ and $1$. Since $X$ is uniformly distributed over this interval, it follows that $$\Pr\left(X\le \frac{y-(-11)}{28}\right)=\frac{y-(-11)}{28},$$ which is what we wanted to prove. Alternately, we have shown that for $y$ between $-11$ and $17$, $F_Y(y)$, the cumulative distribution function of $Y$, is equal to $\frac{y-(-11)}{28}$. Differentiating, we find that between $-11$ and $17$, the density function of $Y$ is $\frac{1}{28}$.

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