# Compute 16 var(x)+32 var(y) for given bivariate CDF

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

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

## 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

• you should check $E[x^2]$ calculation, use wolfram integrator. Jul 13, 2020 at 10:43
• @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. Jul 13, 2020 at 11:00
• How to calculate 2nd moment given CDF? Any clue Jul 13, 2020 at 11:00
• $Y$ 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. Jul 13, 2020 at 11:03
• 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. Jul 13, 2020 at 11:11

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