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fredd [130]
3 years ago
8

. Let X and Y be random variables of possible percent returns (0%, 10%,

Mathematics
1 answer:
bazaltina [42]3 years ago
7 0

(a) The marginal distribution of <em>X</em> is

Pr(<em>X</em> = <em>x</em>) = ∑ Pr(<em>X</em> = <em>x</em>, <em>Y</em> = <em>y</em>)

… = 0.0625 + 0.0625 + 0.0625 + 0.0625

… = 0.25

That is, the first equality follows from the law of total probability, with the sum taken over <em>y</em> from {0, 5, 10, 15}. Each probability Pr(<em>X</em> = <em>x</em>, <em>Y</em> = <em>y</em>) is given in the table to be 0.0625.

Similarly, the marginal distribution of <em>Y</em> is

Pr(<em>Y</em> = <em>y</em>) = 0.25

(b) Yes, they're independent because

Pr(<em>X</em> = <em>x</em>, <em>Y</em> = <em>y</em>) = 0.0625,

and

Pr(<em>X</em> = <em>x</em>) Pr(<em>Y</em> = <em>y</em>) = 0.25 • 0.25 = 0.0625.

(c) The mean of <em>X</em> is

E[<em>X</em>] = ∑ <em>x</em> Pr(<em>X</em> = <em>x</em>)

… = 0.25 ∑ <em>x</em>

<em>… </em>= 0.25 (0 + 5 + 10 + 15)

… = 7.5

and you would find the same mean for <em>Y</em>,

E[<em>Y</em>] = 7.5

The variance of <em>X</em> is

V[<em>X</em>] = E[<em>X</em>^2] - E[<em>X</em>]^2

… = (∑ <em>x</em>^2 Pr(<em>X</em> = <em>x</em>)) - 7.5^2

… = 0.25 (∑ <em>x</em>^2) - 56.25

… = 0.25 (0^2 + 5^2 + 10^2 + 15^2) - 56.25

… = 31.25

and similarly,

V[<em>Y</em>] = 31.25

(each sum is taken with <em>x</em> and <em>y</em> from {0, 5, 10, 15})

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