To find $\text{Var}(Y)$, we can use the law of total variance. The law of total variance states:

[ \text{Var}(Y) = \mathbb{E}[\text{Var}(Y | X)] + \text{Var}(\mathbb{E}[Y | X]) ]

Given:

• $\mathbb{E}(Y | X = x) = 2x$
• $\text{Var}(Y | X = x) = 4x^2$
• $X$ has a uniform distribution on the interval $(0, 1)$, i.e., $X \sim U(0, 1)$

Step 1: Calculate $\mathbb{E}[\text{Var}(Y | X)]$

[ \mathbb{E}[\text{Var}(Y | X)] = \mathbb{E}[4X^2] ]

Since $X \sim U(0, 1)$, the expected value of $X^2$ is:

$\mathbb{E}[X^2] = \int_0^1 x^2 \, dx$

Let's compute this integral:

[ \mathbb{E}[X^2] = \int_0^1 x^2 , dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} ]

So,

[ \mathbb{E}[\text{Var}(Y | X)] = 4 \cdot \mathbb{E}[X^2] = 4 \cdot \frac{1}{3} = \frac{4}{3} ]

Step 2: Calculate $\text{Var}(\mathbb{E}[Y | X])$

$\text{Var}(\mathbb{E}[Y | X]) = \text{Var}(2X)$

Since $\mathbb{E}[Y | X] = 2X$, we calculate the variance of $2X$. Because $X \sim U(0, 1)$, the variance of $X$ is:

[ \text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 ]

We already know $\mathbb{E}[X^2] = \frac{1}{3}$ and $\mathbb{E}[X] = \frac{1}{2}$. Therefore:

$\text{Var}(X) = \frac{1}{3} - \left(\frac{1}{2}\right)^2 = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12}$

Now, since $\text{Var}(2X) = 4 \cdot \text{Var}(X)$:

$\text{Var}(2X) = 4 \cdot \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$

Step 3: Calculate $\text{Var}(Y)$

Finally, we combine these results using the law of total variance:

[ \text{Var}(Y) = \mathbb{E}[\text{Var}(Y | X)] + \text{Var}(\mathbb{E}[Y | X]) = \frac{4}{3} + \frac{1}{3} = \frac{5}{3} ]

So, $\text{Var}(Y) = \frac{5}{3}$.

Would you like more details or have any questions?

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Related Questions:

1. How does the law of total variance apply to other distributions?
2. What if $X$ had a different distribution, such as normal or exponential? How would that change $\text{Var}(Y)$?
3. How do we compute higher moments like $\text{Var}(\text{Var}(Y|X))$?
4. Can $\text{Var}(Y)$ ever be less than $\mathbb{E}[\text{Var}(Y|X)]$?
5. How would the result change if $\text{Var}(Y|X) = \text{constant}$?

Tip: Always double-check your use of expectation and variance identities, especially when dealing with conditional distributions.