We are given the probability density function (PDF) of a random variable $Y$ as:

$f(y) = 1, \quad 0 < y < 1.$

We need to find: (a) The mean ($\mu$), and
(b) The variance ($\sigma^2$) of $Y$.

(a) Mean ($\mu$):

The mean of a continuous random variable is given by:

$\mu = \mathbb{E}[Y] = \int_{-\infty}^\infty y f(y) \, dy.$

Substituting $f(y) = 1$ for $0 < y < 1$:

$\mu = \int_{0}^1 y \cdot 1 \, dy = \int_{0}^1 y \, dy.$

The integral evaluates as follows:

$\mu = \left[ \frac{y^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}.$

So, the mean is:

$\mu = \frac{1}{2}.$

(b) Variance ($\sigma^2$):

The variance of a continuous random variable is given by:

[ \sigma^2 = \mathbb{E}[Y^2] - (\mathbb{E}[Y])^2. ]

Step 1: Compute $\mathbb{E}[Y^2]$:

$\mathbb{E}[Y^2] = \int_{-\infty}^\infty y^2 f(y) \, dy.$

Substituting $f(y) = 1$ for $0 < y < 1$:

$\mathbb{E}[Y^2] = \int_{0}^1 y^2 \cdot 1 \, dy = \int_{0}^1 y^2 \, dy.$

The integral evaluates as follows:

[ \mathbb{E}[Y^2] = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}. ]

Step 2: Compute variance:

[ \sigma^2 = \mathbb{E}[Y^2] - (\mathbb{E}[Y])^2. ]

Substituting $\mathbb{E}[Y^2] = \frac{1}{3}$ and $\mathbb{E}[Y] = \frac{1}{2}$:

$\sigma^2 = \frac{1}{3} - \left(\frac{1}{2}\right)^2 = \frac{1}{3} - \frac{1}{4}.$

Finding the common denominator:

$\sigma^2 = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}.$

So, the variance is:

$\sigma^2 = \frac{1}{12}.$

Final Answers:

(a) Mean ($\mu$) = $\frac{1}{2}$.
(b) Variance ($\sigma^2$) = $\frac{1}{12}$.

Would you like me to clarify any of these steps or expand further?
Here are five related questions to deepen understanding:

1. How is the PDF verified to be valid (integrating to 1)?
2. What is the physical interpretation of mean and variance in this case?
3. How would the calculations change if $f(y)$ was not constant?
4. What are the units of variance compared to the mean?
5. How would the results differ if $y$ had bounds other than $0 < y < 1$?

Tip: Always check that the integral of the PDF over its range equals 1 to confirm it’s a valid probability distribution.