Sophia is vacationing in Monte Carlo On any given night, she takes X dollars to the casino and returns with Y dollars The random variable X has the PDF shown in the figure Conditional on X = x the continuous random variable Y is uniformly distributed between zero and 3 x

Answer 1

Based on your description, we can summarize Sophia's gambling scenario as follows:

1. Random Variable X: This is the amount of money (in dollars) that Sophia takes to the casino on any given night. You mentioned that there is a Probability Density Function (PDF) for X, but you didn't provide the specific PDF. In order to proceed with further analyses, we would need the expression or data defining the PDF of X.

2. Conditional Random Variable Y: Given that Sophia takes \( X = x \) dollars to the casino, the amount she returns, \( Y \), is uniformly distributed between 0 and \( 3x \). Thus, we can denote this as: \[ Y | (X = x) \sim U(0, 3x) \]

Step 1: Finding the PDF of Y

To find the PDF of \( Y \), we start from the conditional distribution of \( Y \) given \( X = x \). We know that:

• The uniform distribution \( U(0, b) \) has a PDF given by: \[ f_Y(y | X = x) = \begin{cases} \frac{1}{b} & \text{if } 0 \leq y \leq b \ 0 & \text{otherwise} \end{cases} \]

For our case, since \( b = 3x \): \[ f_Y(y | X = x) = \begin{cases} \frac{1}{3x} & \text{if } 0 \leq y \leq 3x \ 0 & \text{otherwise} \end{cases} \]

Step 2: Finding the Marginal Probability Density Function of Y

To find the marginal PDF of \( Y \), we can use the law of total probability:

\[ f_Y(y) = \int f_Y(y | X = x) f_X(x) dx \]

This would require knowing the limits of \( x \) for which \( 0 \leq y \leq 3x \). We can solve for \( x \): \[ x \geq \frac{y}{3}. \]

Thus, the integration range for \( x \) will start from \( \frac{y}{3} \) to whatever the upper limit of \( X \) is (let's denote it as \( b \), the maximum value of \( X \)).

Hence, the marginal PDF of \( Y \) becomes:

\[ f_Y(y) = \int_{\frac{y}{3}}^{b} f_Y(y | X = x) f_X(x) dx \]

Substituting \( f_Y(y | X = x) = \frac{1}{3x} \), we obtain:

\[ f_Y(y) = \int_{\frac{y}{3}}^{b} \frac{1}{3x} f_X(x) dx \]

This tells us how to combine the distributions mathematically but requires information about \( f_X(x) \), which we would need to finalize any calculations.

Conclusion

To proceed with numerical computations or specific probability calculations involving \( Y \), the PDF of \( X \) must be provided. Once that is available, we can compute the desired properties of \( Y \) such as its mean, variance, or other specific probabilities.