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Question

X and Y are two continuous random variables with probability density functions $f_{x} (x)$ and $f_{y} (y)$ respectively. If E[•] indicates the expectation operator, then select the correct one of the following relations:

A

E [l n (f_{y} (x))] \leq E [l n (f_{x} (x))]

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B

E [l n (f_{y} (x))] = E [l n (f_{x} (x))]

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C

E [l n (f_{y} (x))] \leq E [l n (f_{x} (y))]

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D

E [l n (f_{y} (x))] \geq E [l n (f_{x} (x))]

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Solution

The correct option is A $E [l n (f_{y} (x))] \leq E [l n (f_{x} (x))]$

Let us assume that, $z = \frac{f_{y} (x)}{f_{x} (x)}$

As $f_{x} (x)$ and $f_{y} (y)$ are probability density functions, $f_{x} (x) \geq 0$ and $f_{y} (x) \geq 0$ . Hence $z \geq 0$ always.
We can use following property of the natural logarithm,
$l n (x) \leq z - 1; f o r z \geq 0$
The equality holds only at z = 1

So, $l n [\frac{f_{x} (x)}{f_{x} (x)}] \leq \frac{f_{y} (x)}{f_{x} (x)} - 1$

$f_{x} (x) l n [\frac{f_{x} (x)}{f_{x} (x)}] \leq f_{y} (y) - f_{x} (x)$

$\int_{- \infty}^{\infty} f x (x) l n [\frac{f_{y} (x)}{f_{x} (x)}] d x \geq \int_{- \infty}^{\infty} f_{y} (x) d x - \int_{- \infty}^{\infty} f_{x} (x) d x$

From the basic properties of probability density function
$\int_{- \infty}^{\infty} f_{x} (u) d u = \int_{- \infty}^{\infty} f_{y} (v) d v = 1$

so, $\int_{- \infty}^{\infty} f_{x} (x) d x l n [\frac{f_{y} (x)}{f_{x} (x)}] d x \leq (1 - 1 = 0)$

$\int_{- \infty}^{\infty} f_{x} (x) l n f_{x} (x) d x - \int_{- \infty}^{\infty} f_{x} (x) l n (f_{x} (x)) d x \leq 0$

$\int_{- \infty}^{\infty} f_{x} (x) l n (f_{y} (x)) d x \leq \int_{- \infty}^{\infty} f_{x} (x) l n (f_{x} (x)) d x$

$E [l n (f_{y} (x))] \leq E [l n (f_{x} (x))]$

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