Uniformly distributed random variable [Solution check]

Question

Mid-term study... Did I do this sample question correct?

Let Y be a random variable that is uniformly distributed on the interval [2,4].

1. What is $f_Y(Y)$?

2. Write an explicit formula for the cumulative distribution function $F_Y(Y)$

3. Find $E(Y)$ and $Var(Y)$

4. A new random variable $W$ is defined by setting $W = aY + b$ where $a$ and $b$ are constants. Find the values of $a$ and $b$ if it is known that $E(W) = -1$ and $Var(W) = 5$.

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So for part 1, I did $f_Y(Y) = \frac{1}{b-a} = \frac{1}{2}$

Part 2: $\int_{2}^{4}{\frac{x-2}{4-2}dx} = 2$

Part 3:

$E(X) = \int_{2}^{4}{x\frac{x-2}{2}dx} = \frac{20}{6}$

$Var(X) = \int_{2}^{4}{x^2\frac{x-2}{2}dx} = 14$

Part 4:

$E(X) = \int_{2}^{4}{(ax\frac{x-2}{2} + bx)dx} = -1 \therefore b = -\left( \frac{1}{6}+ 20a \right)$

$ Var(X) \int_{2}^{4}{(ax^2\frac{x-2}{2} + bx^2)dx} = 5 \therefore a = \frac{5 - 16b}{14}$

Plugging in:

$a = \frac{5 + 16\left( \frac{1}{6} + 20a \right)}{16} = -\frac{53}{198}$

$b = -\frac{1}{6} - 20\left(-\frac{53}{198}\right) = \frac{1027}{198}$

How'd I do?

Answer 1

Problem $1$: It is not quite wrong, but would get downgraded. First, you should call it $f_Y(y)$. Second, you must specify where the density function is $\frac{1}{2}$. This is the interval $[2,4]$. You are probably expected to say it is $0$ elsewhere.

Problem $2$: For $y\lt 2$, we have $F_Y(y)=0$. For $2\le y\le 4$, we have $F_Y(y)=\frac{y-2}{2}$. And finally for $y\gt 4$ we have $F_Y(y)=0$. These follow easily from the fact that $F_Y(y)=\Pr(Y\le y)$. Note that we must specify $F_Y(y)$ for all $y$.

Problem $3$: We want $\int_2^4 y\cdot \frac{1}{2}\,dy$. This is $3$, which is also obvious by symmetry. The wrong density function was being used.

We also want $E(Y^2)-(E(Y))^2$. We already know $E(Y)$. For $E(Y^2)$ find $\int_2^4 y^2\cdot \frac{1}{2}\,dy$. YYou were computing using the wrong density function. Note that the variance is (in general) not $E(Y^2)$.

Problem $4$: We have $E(W)=aE(Y)+b$, and $\text{Var}(W)=a^2\text{Var}(Y)$. Substitute the values for the mean and variance of $Y$ already computed, and solve for $a$ and $b$.

Note there may be two values for $a$ and hence for $b$.