Probability Density Function

[b97704035]

Given f(x) =
a(x + 2) (1<= x <=2)
b(x + 1) (-1 <= x <= 1)
0 (any other condition)

(1) Find the condition of a and b that makes f(x) a probability density function.
(2) Let X be the continuous probability variable of probability density function f(x).
If expectation (＝average) = 2, please find variance V(X).

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I solve the questions as shown on the attach files.
And the answer is
(1) 7a+4b=2 and a>=0, b>=0
(2) V(X) =4/5
Can anyone kindly let me know if it is correct or there is anything I got wrong?

Especially for (2).
While I use another approach to get variance: V(X) = E(X^2)-E(X)^2
with the value of a and b shown in the picture, the answer is different (not 4/5).
Is there something wrong?

 

[Romsek]

You have a problem.

[MATH]f(1) = -\dfrac 1 5 (1+1) = -\dfrac 2 5\\ \text{as $f(x)$ is a pdf it must be that $f(x)\geq 0,~\forall x$}[/MATH]
You need to include this when finding \(\displaystyle a\) and \(\displaystyle b\) and I don't think the problem has a solution as written.

 

[b97704035]

Thanks for your reply.
Now I have known that the value of a and b may be wrong.
May I ask what approach would you use to find V(X) in this question?

Moreoever, how about question(1)? Is the condition correct as I wrote in the picture attached?

Thank you.

You have a problem.

[MATH]f(1) = -\dfrac 1 5 (1+1) = -\dfrac 2 5\\ \text{as $f(x)$ is a pdf it must be that $f(x)\geq 0,~\forall x$}[/MATH]
You need to include this when finding \(\displaystyle a\) and \(\displaystyle b\) and I don't think the problem has a solution as written.

 

[Steven G]

Given f(x) =
a(x + 2) (1<= x <=2)
b(x + 1) (-1 <= x <= 1)

You have one additional information with a and b which you failed to use!

Based on the definition of f(x) which you posted, on one hand f(1) = 3a and on the other hand f(1) =2b. This gives us that 3a=2b. Now you can solve for a and b exactly.