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saul85 [17]
4 years ago
6

Show that the sum of two concave functions is concave. Is the product of two concave functions also concave?

Mathematics
1 answer:
s2008m [1.1K]4 years ago
3 0

Answer:

1.

Let f and g concave functions. Then for a in (0,1), f( (1-\lambda)x + \lambda y ) \geq (1-\lambda)f(x) + \lambda f(y) and g( (1-\lambda)x + \lambda y ) \geq (1-\lambda)g(x) + \lambda g(y)

Now, (f+g)( (1-\lambda)x + \lambda y ) = f( (1-\lambda)x + \lambda y ) + g( (1-\lambda)x + \lambda y )\geq (1-\lambda)f(x) + \lambda f(y) + (1-\lambda)g(x) + \lambda g(y) = (1-\lambda)(f(x)+g(x)) + \lambda (f(x)+g(x))= (1-\lambda)(f+g)(x) + \lambda(f+g)(x)

This shows that the function f+g is concave.

2. Let f and g concave functions.

(fg)( (1-a)x + \lambda y )= f( (1-\lambda)x + \lambda y )g( (1-\lambda)x + \lambda y )  \geq [(1-a)f(x) + \lambda f(y)] [(1-\lambda)g(x) + \lambda g(y)] = (1-\lambda)^2 f(x)g(x) + \lambda (1-\lambda) f(x)g(y)+ \lambda (1-\lambda) f(y)g(x)+ \lambda ^2 f(y)g(y) \neq (1-\lambda)f(x)g(x) + \lambda f(y)g(y)

This shows that the product of two concave functions isn't concave.

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a) 0.20

b) 0.45

c) 0.65

d) Yes

e) Yes

f) Z = X + Y (except when X = 1 and Y = 1)

This is because the successes of X and Y are mutually exclusive events but their failures aren't. X and Y cannot both be 1.

Step-by-step explanation:

Probability of a red set = 20% = 0.20

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e) Is pZ = pX + pY

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Hence, this statement is correct!

f) Z = X + Y

Let's check all the probabilities

when X = 1 and Y = 1, Z = 1

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when X = 0 and Y = 1, Z = 1

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when X = 0 and Y = 0, Z = 0

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