Question
This figure below describes the joint PDF of the random variables X and Y These random variables take values in [0, 2 ]and [0,1] respectively At x = 1 the value of the joint PDF is 1/2
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GPT-4o mini
It seems like you're discussing a joint probability density function (PDF) for two random variables, \(X\) and \(Y\), defined over certain ranges. Since the figure is not available to me, I can help you understand how to analyze a joint PDF based on the information provided.
Given that the joint PDF \(f_{X,Y}(x,y)\) has a value of \(\frac{1}{2}\) at \(x = 1\), we can interpret the implications:
1. **Support of the Joint PDF**: Since the random variables \(X\) and \(Y\) take values in \([0, 2]\) and \([0, 1]\) respectively, the joint PDF is defined in the rectangle defined by these ranges.
2. **Normalization**: To be a valid joint PDF, the total area under the PDF over its support must equal 1. This can be mathematically expressed as:
\[
\int_0^2 \int_0^1 f_{X,Y}(x,y) \, dy \, dx = 1
\]
You'll want to check if the function is normalized over the entire support range.
3. **Value of the PDF**: When \(f_{X,Y}(1,y) = \frac{1}{2}\), this provides information about the distribution at that specific point in the \(x\)-domain. You would need to understand how this function behaves for other values of \(x\) in order to get the complete picture of the joint distribution.
4. **Marginal Distributions**: You can find the marginal PDF of \(X\) and \(Y\) by integrating:
- Marginal of \(X\):
\[
f_X(x) = \int_0^1 f_{X,Y}(x,y) \, dy \quad \text{for } 0 \leq x \leq 2
\]
- Marginal of \(Y\):
\[
f_Y(y) = \int_0^2 f_{X,Y}(x,y) \, dx \quad \text{for } 0 \leq y \leq 1
\]
5. **Conditional Distributions**: If you are interested in \(Y\) given \(X = 1\), you can find the conditional PDF:
\[
f_{Y|X}(y|1) = \frac{f_{X,Y}(1,y)}{f_X(1)}
\]
If you have any specific questions or need help with computations, feel free to provide more details!
Given that the joint PDF \(f_{X,Y}(x,y)\) has a value of \(\frac{1}{2}\) at \(x = 1\), we can interpret the implications:
1. **Support of the Joint PDF**: Since the random variables \(X\) and \(Y\) take values in \([0, 2]\) and \([0, 1]\) respectively, the joint PDF is defined in the rectangle defined by these ranges.
2. **Normalization**: To be a valid joint PDF, the total area under the PDF over its support must equal 1. This can be mathematically expressed as:
\[
\int_0^2 \int_0^1 f_{X,Y}(x,y) \, dy \, dx = 1
\]
You'll want to check if the function is normalized over the entire support range.
3. **Value of the PDF**: When \(f_{X,Y}(1,y) = \frac{1}{2}\), this provides information about the distribution at that specific point in the \(x\)-domain. You would need to understand how this function behaves for other values of \(x\) in order to get the complete picture of the joint distribution.
4. **Marginal Distributions**: You can find the marginal PDF of \(X\) and \(Y\) by integrating:
- Marginal of \(X\):
\[
f_X(x) = \int_0^1 f_{X,Y}(x,y) \, dy \quad \text{for } 0 \leq x \leq 2
\]
- Marginal of \(Y\):
\[
f_Y(y) = \int_0^2 f_{X,Y}(x,y) \, dx \quad \text{for } 0 \leq y \leq 1
\]
5. **Conditional Distributions**: If you are interested in \(Y\) given \(X = 1\), you can find the conditional PDF:
\[
f_{Y|X}(y|1) = \frac{f_{X,Y}(1,y)}{f_X(1)}
\]
If you have any specific questions or need help with computations, feel free to provide more details!
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