Uniform distribution of $Z$ given $Y\sim U(2, 5)$

Question

Given that $Y \sim U(2, 5)$ and $Z = 3Y - 4$, what is the distribution for $Z$?

I've worked out that for $Y \sim N(2, 5)$, $Z \sim N(2, 45)$ since

$$\mu=3\cdot2 - 4 = 2$$

and

$$\sigma^2=3^2 \cdot 5 = 45$$

I'm wondering how the working differs when we have a uniform distribution, rather than a normal distribution?

Sorry if a similar question has been asked before - I could not find anything on my search!

Thanks!

Answer 1

Call $U$ the CDF of $U(2,5)$, it is $U(t) = \frac{1}{3}t - \frac{2}{3}$.

$A(t) = \mathbb P(Z \leq t) = \mathbb P (3Y-4 \leq t) = \mathbb P (Y\leq \frac{t+4}{3}) = U(\frac{t+4}{3}) = \frac{1}{3}[\frac{t+4}{3}] -\frac{2}{3}$ for $2 \leq \frac{t+4}{3} \leq 5$, $0$ elsewhere, i.e. for $2 \leq t \leq 11$.

Thus $a(t) = A'(t) = \frac{1}{9}$ in $2 \leq t \leq 11$, $0$ elsewhere is a uniform distribution.

Answer 2

A general technique for this type of problems is to find the CDF for $Y$, then find CDF for $Z$ accordingly, and recognize what that CDF is.

However, for uniform distribution, affine transformation of uniform distribution is still uniform as well. So the lowest value of $Z$ is 2, and the highest value is 11, so $Z$ is uniform from 2 to 11.

If you want to know why that claim is true (affine transformation of uniform distribution is also affine), consider the following scenarios:

1. Transforming $Y$ to $3Y$. If $Y$ is uniform then the PDF of $Y$ is a straight line (a ramp function). So it's zero until 2, then ramps up to 1 at 5, then stays 1 after 5. The transformation from $Y$ to $3Y$ merely scales the x-axis, so it's still a ramp function, and that's a CDF of a uniform distribution

2. Transforming from $3Y$ to $3Y-4$, this is just moving the "location" of the variables to the left by 4, so it does not affect the "shape" of the distribution

Answer 3

If $Y\sim \mathsf {Unif}(2,5)$ and $Z=3Y-4$ then $Z\sim \mathsf {Unif}(2,11)$.

The transformation stretches the distribution of $Y$ by a factor of $3$ and then shifts it $4$ units to the left. Recalling that

$$F_Y(y) = \mathsf{P}({Y\leq y})=\frac{y-2}{3}$$ for $2< y <5$ we get that for $2< z <11$,

$$\begin{align*} F_Z(z) &=\mathsf P({Z\leq z})\\\\ &=\mathsf P({3Y-4\leq z})\\\\ &=\mathsf P({3Y\leq z+4})\\\\ &=\mathsf P\left(Y\leq \frac{z+4}{3}\right)\\\\ &=F_Y\left(\frac{z+4}{3}\right)\\\\ &=\frac{z-2}{9} \end{align*}$$

Because the density function of a random variable is the derivative of its CDF, we see that for $2< z <11$, the density function of $Z$ is $$f_Z(z) = \frac{1}{9}$$

Answer 4

Let $a,b\in\mathbb R$ with $a<b$, and let $U$ be uniformly distributed over $(0,1)$.

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In general a random variable $X$ is uniformly distributed over interval $(a,b)$ if and only if its CDF can be prescribed by:

• $x\mapsto0$ if $x\leq a$
• $x\mapsto\frac{x-a}{b-a}$ if $a<x\leq b$
• $x\mapsto1$ otherwise.

Now observe that on base of this it can easily be deduced that $a+(b-a)U$ is uniformly distributed over interval $(a,b)$.

Actually we can write $X=a+(b-a)V$ where $V=\frac{X-a}{b-a}$ and where $V$ is uniformly distributed over $(0,1)$.

Applying this on your question we conclude that $Y=2+3V$ where $V$ is uniformly distributed over $(0,1)$ and consequently $$Z=3Y-4=3[2+3V]-4=2+9V$$allowing the conclusion that $Z$ is uniformly distributed over $(2,11)$.