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\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

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    $\begingroup$ you should check $E[x^2]$ calculation, use wolfram integrator. $\endgroup$
    – gunes
    Jul 13, 2020 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. $\endgroup$ Jul 13, 2020 at 11:00
  • $\begingroup$ How to calculate 2nd moment given CDF? Any clue $\endgroup$ Jul 13, 2020 at 11:00
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    $\begingroup$ $Y$ 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, 2020 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. $\endgroup$ Jul 13, 2020 at 11:11

1 Answer 1

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... 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. $$f(x)=\int f(x,y)dy$$

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  • $\begingroup$ sir i have attempted after your hint. Pls check my above edit $\endgroup$ Jul 13, 2020 at 10:35
  • $\begingroup$ ......................... $\endgroup$ Jul 13, 2020 at 10:36

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