Question

Assume the length X, in minutes, of a particular type of telephone conversation is a random variable with pdf f(x)=.2e^(-.2x), x>0. Find E[e^(.1X)+3X].
I started this problem by finding E(X)=integral x f(x) dx and with integration by parts and u substitution found the E(X) to be 5.
I don't know how to proceed to find E[e^(.1X)+3X].

Answer 1

E(x) = integral x * f(x) dx

Answer 2

Thanks @perl. I actually put that in my question. I found E(X) I just don't know how to proceed to find E[e^(.1X)+3X].

Answer 3

E[e^(.1X)+3X]=E[e^(.1X)]+3E[X]

Answer 4

\[E[g(X)]=\int\limits_{\mathbb{R}}g(x)f(x)dx\]

Answer 5

we can use Zarkon's suggestion,
A useful formula :
E[aX + b] = aE[X] + b

Answer 6

So \[E[e ^{.1X}+3X] = 3(5)+e ^{.1X}\] ?

Answer 7

$$\Large{ f(x)=.2e^{-.2x}\\
E[e^{.1X}+3X]=E[e^{.1X}]+3E[X]\\
\\ \therefore\\

= \int_{0}^{\infty}e^{.1x}(0.2e^{-.2x})dx + 3\int_{0}^{\infty}x(0.2e^{-.2x})dx
\\ = \int_{0}^{\infty}0.2e^{-.1x}dx + 3(5)

}
$$

Answer 8

@Zarkon
why is
$$ \large E[g(X)]=\int_{-\infty }^{\infty} g(x)f(x)dx

$$
not
$$ \large E[g(X)]=\int_{-\infty }^{\infty} g(x)f(g(x))dx

$$
or are they equal

Answer 9

it might be easier to show with a discrete p.m.f

Answer 10

http://mathworld.wolfram.com/ExpectationValue.html

Answer 11

those two are most definitely not equal

Answer 12

in general

Answer 13

i guess im trying to replace g(x) consistently with x in the formula above.
but the underlying random variable X is still the same, is that whats going on

Answer 14

Thank you @zarkon and @perl for your comments.
Can you tell me if in fact this should be \[=\int\limits_{0}^{\infty}e ^{.1X}(.2e ^{-.2X})dx + 3 \int\limits_{0}^{\infty}x(.2e ^{-.2x})dx\]
and therefore = 2+ 3(5) = 17

Answer 15

yes thats correct

Answer 16

thank you!

Answer 17

what if you did
Y = g(X)
then
E [g(X)] = E[Y] = integral y * f(y)

Answer 18

\[E(g(X)]=\int\limits_{\mathbb{R}}g(x)f_{X}(x)dx\]

\[Y=g(X)\]

\[E(Y)=\int\limits_{\mathbb{R}}yf_{Y}(y)dy\]


you can prove that they are equal

Answer 19

f_X and f_Y must be different functions.
f_X is the given probability density function

since you defined
X = g(Y)
then
Y = g^-1 (X)

I will work on this more later . thanks :)

Answer 20

you have to be a little careful...g might not be one-to-one