Calculate E(g(X)) for Random Variable X with E(X)=6.2, Var(X)=0.8

[das1]

This problem:

A random variable X has expected value E(X) = 6.2 and variance Var(X) = 0.8. Calculate the expected value of g(X) where g(x) = 7x + 2.

Do I just plug in numbers here? I've never seen this kind of problem before.

 

[Jameson]

Here's a useful fact: if $f(x)$ is the pdf of a continuous random variable then \(\displaystyle E[g(x)]=\int_{-\infty}^{\infty}g(x)f(x)dx\)

Substitute in $g(x)$ and use rules for integration to try to see what you can do.

 

[das1]

E(7x+2) = ∫(7x+2)*f(x) dx
I'm still stuck at integrating (7x + 2)*f(x). If we don't know what f(x) is, I don't know what to integrate. When you refer to the rules of integration, is there a certain rule you are referring to? Is there a particular way to write ∫f(x) dx when you don't know what f(x) is?

 

[Jameson]

Not in general, but in this case we can use some tricks.

Multiply through and you get:

\(\displaystyle \int_{-\infty}^{\infty}7x \cdot f(x)+2 \cdot f(x)dx=\int_{-\infty}^{\infty}7x \cdot f(x)dx+\int_{-\infty}^{\infty} 2 \cdot f(x)dx=7\int_{-\infty}^{\infty}x \cdot f(x)dx+2\int_{-\infty}^{\infty}f(x)dx\)

Now use some definitions that you have been given to proceed. :)

 

[das1]

OK right, I recognize ∫x*f(x) dx as the mean, so can I just plug in 6.2 for that? But what about 2*∫f(x) dx? I suspect that's related to the variance but nothing jumps out at me.

 

[Jameson]

If $f(x)$ is a proper pdf for a continuous random variable, then by definition \(\displaystyle \int_{-\infty}^{\infty}f(x)=1\).

We kind of did this the long way, but it's good for understanding. It's also a property that: \(\displaystyle E[aX+b]=E[aX]+E=aE[X]+b\), where $X$ is a continuous random variable and $a,b$ are real numbers.

 

[Jameson]

Yes. Here is a page that covers the basics for discrete and continuous distributions.

 

[das1]

OK thank you, that clears it up.
So the answer would just be 7*6.2+2*1 = 45.4?