2

There are three different incomes, x, and their proportions, f (x).
$ 10,000 0.40

$ 30,000 0.40

$ 50,000 0.20

How do I calculate the joint distribution for X1 and X2, which are a random sample of two incomes?

I'm new to this, so any help is appreciated. I feel like the proportions are important here, but I don't think I'd multiply the income by the proportion.

Rosie E
  • 55
  • 8
  • Assuming only these 3 incomes are possible, it seems like you just need to do the obvious: $P (X_1=i, X_2=j)= P (X_1=i) P (X_2=j)$ – ved Jun 30 '14 at 03:31

1 Answers1

1

You will have to make some assumptions. Probably you are expected to assume that the random variables $X_1$ and $X_2$ are independent. Under that assumption, which you should state explicitly, we have $\Pr(X_1=a \land X_2=b)=\Pr(X_1=a)\Pr(X_2=b)$.

Now we just need to compute. For example, $\Pr(X_1=10000\cap X_2=10000)=(0.4)(0.4)$, $\Pr(X_1=10000\cap X_2=30000)=(0.4)(0.4)$, and $\Pr(X_1=10000\cap X_2=50000)=(0.4)(0.2)$. There are $6$ other entries.

Your notation might be different. You may be expected to write $f_{X_1,X_2}(10000,10000)=(0.4)(0.4)$, with $8$ other similar expressions.

André Nicolas
  • 507,029
  • Ok this makes sense. I need to also find the sample mean, which in this case would be 1/2 times X1+X2+….+Xn. Which values am I supposed to use because it seems like my answer should be in the income, not the probability. – Rosie E Jun 30 '14 at 03:46
  • 1
    The sample mean is $\frac{X_1+X_2}{2}$. We want the expectation of this. That's $\frac{1}{2}(E(X_1)+E(X_2))$, i.e. it is the same as the mean of $X_1$ (or $X_2$). No big surprise!. Now to find $E(X_1)$, which is our answer, just calculate $(10000)(0.4)+(30000)(0.4)+(50000)(0.2)$. – André Nicolas Jun 30 '14 at 03:54
  • So the expected value of X1 is 26,000 and the sample mean would be the same thing because the expected value for X1 and X2 is identical? Is this correct? So X1 doesn't have to be one of the 3 possible incomes? – Rosie E Jun 30 '14 at 04:06
  • 1
    The random variable $X_1$ takes on the values $10000$, $30000$, $50000$ with the given probabilities. If we pick a person at random from the population to be our first sample, we do not know in advance what the income will be, that's why it is a random variable. – André Nicolas Jun 30 '14 at 04:14