Find E(X), Var(X), SD(X)

[Aurelius032]

Find E(X), Var(X) Standard Deviation(X) if a random variable X is given by its density function f(X) such that:

f(x)=0, if x≤0
f(x)=6(x-x^2), if 0<x≤1
f(x)=0, if x>1

My work:

Please see attached.

This is what I have.. I got E(X) = 0 but I do not think this is right. Can anyone check to see if I am doing this wrong? If so, can someone give me hints as to how I can solve this problem?

Also, How do you find VAR(X) and SD(X)??

 

[Ishuda]

Find E(X), Var(X) Standard Deviation(X) if a random variable X is given by its density function f(X) such that:

f(x)=0, if x≤0
f(x)=6(x-x^2), if 0<x≤1
f(x)=0, if x>1

My work:

Please see attached.

This is what I have.. I got E(X) = 0 but I do not think this is right. Can anyone check to see if I am doing this wrong? If so, can someone give me hints as to how I can solve this problem?

Also, How do you find VAR(X) and SD(X)??

The expected value of g(x) with the probability distribution f(x) is given by, in this case,
E[g(x)] = \(\displaystyle 6 \int_0^1 g(x) (x - x^2) dx\)
The function does not go outside the integration. So, for the expected value of x, what is
E(x) = \(\displaystyle 6 \int_0^1 x (x - x^2) dx\)
and use the definitions for the variance and standard deviation, see
http://en.wikipedia.org/wiki/Variance
for example

 

[Aurelius032]

The expected value of g(x) with the probability distribution f(x) is given by, in this case,
E[g(x)] = \(\displaystyle 6 \int_0^1 g(x) (x - x^2) dx\)
The function does not go outside the integration. So, for the expected value of x, what is
E(x) = \(\displaystyle 6 \int_0^1 x (x - x^2) dx\)
and use the definitions for the variance and standard deviation, see
http://en.wikipedia.org/wiki/Variance
for example

Ishuda - I think I got it. Can you confirm that this is correct?

 

[Ishuda]

Ishuda - I think I got it. Can you confirm that this is correct?

Be careful with your equal signs. For example
\(\displaystyle 6 \int_0^1 x (x - x^2) dx \)
is not equal to 6 (x-x²) although I think the rest of that shows what you really meant. The expected value (mean) is \(\displaystyle \frac{1}{2}\)

For the variance, you have once again put in the wrong values under the integral although the formula is correct:
Var(x) = \(\displaystyle E(x^2) - E^2(x) = 6 \int_0^1 x^2 f(x) dx - E^2(x)\)
The integral for E(x²) should be
E(x²) = \(\displaystyle 6 \int_0^1 x^2 f(x) dx = 6 \int_0^1 x^2 (x - x^2) dx\)

BTW: Since the integral is from 0 to 1, you don't have to take the limits, you can just evaluate it at 1. If you were evaluating just the function f(x) itself at x equal 1, you would have to take the limit from the proper side to get the correct answer.

 

[Aurelius032]

Be careful with your equal signs. For example
\(\displaystyle 6 \int_0^1 x (x - x^2) dx \)
is not equal to 6 (x-x²) although I think the rest of that shows what you really meant. The expected value (mean) is \(\displaystyle \frac{1}{2}\)

For the variance, you have once again put in the wrong values under the integral although the formula is correct:
Var(x) = \(\displaystyle E(x^2) - E^2(x) = 6 \int_0^1 x^2 f(x) dx - E^2(x)\)
The integral for E(x²) should be
E(x²) = \(\displaystyle 6 \int_0^1 x^2 f(x) dx = 6 \int_0^1 x^2 (x - x^2) dx\)

BTW: Since the integral is from 0 to 1, you don't have to take the limits, you can just evaluate it at 1. If you were evaluating just the function f(x) itself at x equal 1, you would have to take the limit from the proper side to get the correct answer.

Thanks for the insight. How does this look?

 

[Ishuda]

Thanks for the insight. How does this look?

The value for E(x²) doesn't appear to be correct:
E(x²)=\(\displaystyle 6 \int_0^1 x^2 f(x) dx = 6 \int_0^1 x^2 (x - x^2) dx =
6 [ \int_0^1 x^3 dx - \int_0^1 x^4 dx ]\)