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Natasha2012 [34]
4 years ago
8

Help ASAP!!!! Please , 15 points

Mathematics
2 answers:
Feliz [49]4 years ago
8 0

Answer:

im probably wrong but i think it's C

Allisa [31]4 years ago
6 0
The answer is B (37 degrees)
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Let X have the uniform distribution U(0, 2) and let the conditional distribution of Y , given that X = x, be U(0, x). Find the j
Elina [12.6K]

Answer:

f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x

E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \

Step-by-step explanation:

We have two random variables X and Y. X \sim Unif(0,2) and given that X=x, Y has uniform distribution (0,x)

From the definition of the uniform distribution we have the densities for each random variable given by:

f_X (x) =\frac{1}{2} , 0\leq x\leq 2

f_{Y|X} (y|x) = \frac{1}{x}, 0\leq y \leq x

And on this case we can find the joint density with the following formula:

f(x,y) = f_{Y|X}(y|x) f_X (x)

And multiplying the densities we got this:

f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x

Now with the joint density we can find the expected value E(Y|x) with the following formula:

E(Y|x) = \int y f_{Y|X}(y|x)dx

And replacing we got:

E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \

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Answer:

The answer is 21.

Step-by-step explanation:

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