The problem you provided is to calculate the expected value $E(Y)$ of a continuous random variable with the probability density function (PDF) given by $f_Y(y) = \frac{1}{4} e^{-y/4}$, where $y$ ranges from 0 to $\infty$.

The expected value $E(Y)$ is calculated by the following formula:

$E(Y) = \int_0^\infty y f_Y(y) \, dy$

Substituting the given PDF into the formula:

$E(Y) = \int_0^\infty y \cdot \frac{1}{4} e^{-y/4} \, dy$

Step 1: Simplifying the integral

First, we can take the constant $\frac{1}{4}$ outside the integral:

$E(Y) = \frac{1}{4} \int_0^\infty y e^{-y/4} \, dy$

Step 2: Solve the integral

To solve the integral, use the substitution method. Let:

$u = \frac{y}{4}, \quad \text{so} \quad du = \frac{1}{4} dy \quad \Rightarrow \quad dy = 4 du$

Now, rewrite the integral in terms of $u$:

$E(Y) = \frac{1}{4} \int_0^\infty 4u \cdot 4 e^{-u} \, du = \frac{1}{4} \cdot 16 \int_0^\infty u e^{-u} \, du$

Thus, we now need to compute:

$\int_0^\infty u e^{-u} \, du$

This is a standard integral and is known to be equal to 1 (it’s the Gamma function $\Gamma(2) = 1! = 1$).

Step 3: Final Answer

Therefore, the expected value becomes:

$E(Y) = \frac{1}{4} \cdot 16 \cdot 1 = 4$

So, the expected value $E(Y) = 4$.

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Would you like a more detailed explanation of the steps? Let me know if you have any questions!

Additional Questions for Practice:

1. What would the expected value be if the PDF were $f_Y(y) = \frac{1}{5} e^{-y/5}$?
2. How would you calculate the variance of a continuous random variable with an exponential distribution?
3. What is the general formula for the expected value of an exponential distribution with rate $\lambda$?
4. How does the shape of the exponential distribution affect the expected value?
5. What happens to the expected value if the exponential distribution is truncated at a certain value?

Tip: When working with exponential distributions, remember that the expected value is always the inverse of the rate parameter $\lambda$.