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Zolol [24]
3 years ago
14

(1 point) Suppose that for two random variables X and Y the joint density function is f(x,y)=6xe−x(y+6), for x>0 and y>0.

Find each of the following. (a) fX|Y(x,y)= (b) fY|X(x,y)=
Mathematics
1 answer:
andriy [413]3 years ago
3 0

First find the marginal distributions:

f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^\infty6e^{-x(y+6)}\,\mathrm dy

\implies f_X(x)=\begin{cases}6e^{-6x}&\text{for }x>0\\0&\text{otherwise}\end{cases}

f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\int_0^\infty6e^{-x(y+6)}\,\mathrm dx

\implies f_Y(y)=\begin{cases}\dfrac6{(y+6)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}

Then to get each conditional distribution, divide the joint distribution by the corresponding marginal distribution.

a.

f_{X|Y}(x,y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}x(y+6)^2e^{-x(y+6)}&\text{for }x>0\\0&\text{otherwise}\end{cases}

b.

f_{Y|X}(x,y)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}xe^{-xy}&\text{for }y>0\\0&\text{otherwise}\end{cases}

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Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

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