Thread: [SOLVED] mean and variance

Let X1,X2,X3,X4 be four iid random variables having the same pdf f(x)=2x, 0<x<1, zero elsewhere. Find the mean and variance of the sum Y of these four random variables. ??

Originally Posted by ninano1205

Let X1,X2,X3,X4 be four iid random variables having the same pdf f(x)=2x, 0<x<1, zero elsewhere. Find the mean and variance of the sum Y of these four random variables. ??

I think you mean to say that $\displaystyle Y=X_1+X_2+X_3+X_4$?

Thus, $\displaystyle E\left[Y\right]=E\left[X_1+X_2+X_3+X_4\right]=E\left[X_1\right]+E\left[X_2\right]+E\left[X_3\right]+E\left[X_4\right]$

Since they all have the same pdf, we can say that $\displaystyle E\left[Y\right]=4\int_0^1 \left[x\cdot\left(2x\right)\right]\,dx=8\int_0^1 x^2\,dx$. Can you take it from here?

For variances, $\displaystyle \text{Var}\left[Y\right]=\text{Var}\left[X_1+X_2+X_3+X_4\right]=\text{Var}\left[X_1\right]+\text{Var}\left[X_2\right]+\text{Var}\left[X_3\right]+\text{Var}\left[X_4\right]$

Since they all have the same pdf, we can say that $\displaystyle \text{Var}\left[Y\right]=4\int_0^1\left[x^2\cdot\left(2x\right)\right]\,dx=8\int_0^1 x^3\,dx$

Can you take it from here? Does this make sense?

I got the mean = 8/3 which is right but the variance doesnt seem to be right. Following ur way the var comes out to be 2 but the answer should be 2/9. What's wrong with it?

Originally Posted by ninano1205

I got the mean = 8/3 which is right but the variance doesnt seem to be right. Following ur way the var comes out to be 2 but the answer should be 2/9. What's wrong with it?

Woops...I forgot something...

$\displaystyle Var\left[x\right]=E\left[x^2\right]-{\color{red}\left(E\left[x\right]\right)^2}$

So you would need to evaluate $\displaystyle 4\left[\int_0^1 2x^3\,dx-\left(\int_0^1 2x^2\,dx\right)^2\right]$ to get the variance (I multiplied the value by 4 because each r.v. has the same variance). It should be $\displaystyle \frac{2}{9}$

Does this make sense?

That's right.
You found the second moment and not the variance.