1
$\begingroup$

\begin{equation} {F(x,y)} = \begin{cases} 0 & \text{if $x<0$ or $y<0 $} \\ \frac{1-e^{-x}}{4} & \text{if $x>0, 0 \leq y <1$} \\ 1-e^{-x}& \text{if $x \geq 0, y \geq1$} & \text{} \end{cases} \end{equation}

My attempt:

F(x,y) = F(x)F(y) , where F(y)=1. So can i say x and y are independent? But here F(x) changes wrto y domain right?

To calculate $E(x) = \int_{x=0}^{\infty} f_X(x) dx \\ f_X(x) = \int_{x=0}^{\infty} \int_{y=0}^{1} \frac{1-e^{-x}}{4} dx dy + \int_{x=0}^{\infty} \int_{y=1}^{\infty} {1-e^{-x}} dx dy $

But then second integral is going to infinity due to y between 1 to infinity.

How to solve this - var(x), var(y)?

Another attempt:

pdf - f(x,y) = differentiate F(x,y) wrto x and y

then i am getting f(x,y)=0 as there is no y-variable

another doubt -

But as F(x,y)=F(x)*1 (F(y)=1) --> x and y are independent random variables

THen i can i directly differentiate F(x,y)=F(x) wrto x to find f(x) and f(y)=0 as F(y) = 1

I am confused. Please enlighten


My attempt after getting hints from 1st answer:

$f(x)=e^{-x}$ (on differentiating F(x) wrto x)

$E(x)=\int_{x=0}^{\infty} xe^{-x} dx = 1$

$E(x^2)=\int_{x=0}^{\infty} x^2e^{-x} dx = 2$

$var(X)=2-1=1$ ---> (is this correct?)

$f(y)=0 $(on differentiating $F_Y(y)$ ---> is this correct???

So $E(y)=E(y^2)=0$

But again,

$E(y)=\int_{y=0}^{\infty}(1-F_Y(y))dy$ ---> expectation interms of cdf

$E(y)=\int_{y=0}^{1}(1-\frac{1}{4})dy+\int_{y=1}^{\infty}(1-1)dy$

But here E(y)=3/4. But if i calculate using pdf $f_Y(y)$, i am getting 0.

Pls clarify. Answer given in textbook is 22. I recently started studying prob and statistics on my own. kindly help

$\endgroup$
7
  • 1
    $\begingroup$ you should check $E[x^2]$ calculation, use wolfram integrator. $\endgroup$ – gunes Jul 13 '20 at 10:43
  • $\begingroup$ @gunes sir corrected it. Now i am getting var(x)=1. But to find pdf of y random variable, is it wrong to differentiate $F_Y(y)$? if so, then f(y)=0. $\endgroup$ – Nascimento de Cos Jul 13 '20 at 11:00
  • $\begingroup$ How to calculate 2nd moment given CDF? Any clue $\endgroup$ – Nascimento de Cos Jul 13 '20 at 11:00
  • 1
    $\begingroup$ $Y$ is a discrete RV, don't take the derivative. $P(Y=0)=1/4$ and $P(Y=1)=3/4$ according to the jumps in CDF. $\endgroup$ – gunes Jul 13 '20 at 11:03
  • $\begingroup$ understood now. Yes it is discrete random variable. It did not click me. In fact it is hybrid function of continious and discreet random vector. Now I am getting E(y)=0.75, E(y^2)=0.75 and var(y)=0.1875. Now 16*var(x)+32*var(y)=16*1+32*0.1875=22. Thank you for patiently explaining me things sir. $\endgroup$ – Nascimento de Cos Jul 13 '20 at 11:11
2
$\begingroup$

... But here F(x) changes wrto y domain right?

This is why you cannot write $F(x,y)=F(x)F(y)$ in the above expression in the way you do. If $F(y)=1$, then $F(x)=F(x,y)$ but still $y$ is in the conditions, however $F(x)$ is a function of $x$ only.

But, if I'm not mistaken you can factorize it as set $F(y)=\begin{cases}1/4 &0\leq y<1\\1& y\geq 1\\0&\text{else}\end{cases}$, and $F(x)=1-e^{-x}$ for $x\geq 0$.

In your integral for finding $f(x)$, you should be using $f(x,y)$, not the joint CDF, and you should be integrating wrt $y$, i.e. $$f(x)=\int f(x,y)dy$$

$\endgroup$
2
  • $\begingroup$ sir i have attempted after your hint. Pls check my above edit $\endgroup$ – Nascimento de Cos Jul 13 '20 at 10:35
  • $\begingroup$ ......................... $\endgroup$ – Nascimento de Cos Jul 13 '20 at 10:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.