# Random variables - density functions

• January 14th 2009, 06:11 PM
GenericNameA
Random variables - density functions
(a) If Z1 and Z2 are independent and have N(0,1) densities, let

X1 = a11Z1 + a12Z2 + c1 and let X2 = a21Z1 + a22Z2 + c2

where a11,a12,a21,a22,c1 and c2 are constants. Determine the joint density of X1 and X2.

(b) Suppose that the random variable X on (0,1) has density

f(x) = 3x2 0 < x < 1

Determine the density functions of

(i) U = X1/2
(ii) V = -log(X)

(c) Now suppose Y is uniformly distributed on (0,1) and independent of X. Determine the density of

W = max(X,Y)
• January 14th 2009, 06:53 PM
Last_Singularity
Quote:

Originally Posted by GenericNameA
(a) If Z1 and Z2 are independent and have N(0,1) densities, let

X1 = a11Z1 + a12Z2 + c1 and let X2 = a21Z1 + a22Z2 + c2

where a11,a12,a21,a22,c1 and c2 are constants. Determine the joint density of X1 and X2.

(b) Suppose that the random variable X on (0,1) has density

f(x) = 3x2 0 < x < 1

Determine the density functions of

(i) U = X1/2
(ii) V = -log(X)

(c) Now suppose Y is uniformly distributed on (0,1) and independent of X. Determine the density of

W = max(X,Y)

Part B (i):
We are given that $f_X(x) = 3X^2, X \in [0,1]$. And given that $U = X^{1/2}$, we need the distribution of $U$

Step 1: Determine the domain of the random variable $U$. Since $X$ is at most $1$, $U$ is at most $1$. And similar, because $X$ is at least $0$, $U$ is at least $0$.
So $U \in [0,1]$ as well.

Step 2: find the CDF of $U$: $F_U(u)$
$= P(U < u)$

$= P(X^{1/2} < u)$

$=P(X < u^2)$

$=F_X(u^2)$

Step 3: find the pdf of $U$
And since $f_U(u) = \frac{\partial F_U}{\partial U}$ and we concluded that $F_U(u) = F_X(u^2)$, it must be the case that:

$f_U(u) = \frac{\partial F_X}{\partial U}$

$= f_x(u^2) \frac{\partial (u^2)}{\partial u}$

$=2u f_x(u^2)$

$=2u 3 (u^2)^2$

$=6u^5$

Step 4: Vary that indeed this is a pdf:
$\int_0^1 6u^5 du = 1$ as desired.

Part B (ii):
This one is a bit tougher. But we can proceed as before:

Step 1: The boundaries of $X$ are $0,1$. So if we let $V = - log(x)$, then the boundaries of $V$ become $0$ and positive infinity.
So $V \in [0, \infty]$

Step 2: find the CDF of $V$: $F_V(v)$
$= P(V < v)$

$= P(-log(X) < v)$

$= P(10^{-log(X)} < 10^v)$

$= P(1/X < 10^v)$

$= P(X < 10^{-v})$

$= F_X(10^{-v})$

Step 3: find the PDF of $V$: $f_V(v)$
$= \frac{\partial F_X(10^{-v})}{\partial v}$

$= -ln(10)(10^{-v}) f_x(10^{-v})$

$=3 ln(10) (1/100)^v (10^{-v})$

Step 4: verify that this is indeed a valid pdf:
$\int_0^\infty 3 ln(10) (1/100)^v (10^{-v}) dv = 1$ as desired.

Part C:
We proceed as before by finding the cdf of $W$:
$= P(W < w)$

Now, here's the trick. Since we have defined $W = max(X,Y)$ or that $W$ represents the larger of $X$ and $Y$, it means that whenever $W$ is less than a constant $w$, then it must be the case that both $X$ and $Y$ are less than that constant $w$. So we can determine that:
$= P(W < w)$
$= P(X < w, Y < w)$
$= P(X < w) P(Y < w)$ (by independent of $X$ and $Y$)
$= F_X(w) F_Y(w)$

We know that $F_X(x) = \int f_X(x) dx = \int 3X^2 dx= X^3$ and that $F_Y(y) = \int f_Y(y) dy = \int 1 dy = Y$

So $F_X(w) F_Y(w) = (W^3)(W) = W^4$

We conclude that the cdf of $W$ is: $F_W(w) = W^4$

Then, of course, the pdf of $W$ is: $f_W(w) = 4W^3, W \in [0,1]$

Verify that this is a valid pdf and you are done.

Part A:
I am too tired lol...just scroll to the section that says "bivariate case" at: http://en.wikipedia.org/wiki/Bivariate_normal

Plug in the appropriate variables into the formula...