Good evening sir, I m glad to watch ur lectures and learned a lot. It is a desire of mine to meet u someday . I hope we will meet someday and discuss on stochastic. Regards
+Merih LEBLEBİCİ Yes, you are right. That was a slip on my part. When the random variables X and Y are independent, the pdf of Z =X-Y is given by fz(z) = fx(-z) * fy(z) .
First, draw a line for X-Y = -1. You will see that when x=0, y=1, or the line intercepts the Y-axis on the positive side as shown there. Also when y=0, x=-1 and so the line intercepts the X-axis on the negative side. This is true for any z
This video was very helpful. I am still a little stuck on one thing: Just like in the second example, I have two random variables that are both greater than or equal to zero (X>0, Y>0). I solved for the joint density such that f(z)= integral fx(x)fy(x-z) dx. I solved the integral but I still don't know what limits to evaluate it at. Through trial and error I found that for z0, Y>0) such that the f(z)= integral fx(x)fy(x-z) dx?
+Aaron Hendrickson. If you watch the video from T=4.16 min onward, you can see the non-negative case (X > 0, Y > 0) done separately. At T= 11.05 min, you can see the limits of integration for z > 0 and z < 0. By the way you should do the x integral first and y integral second as it is easier for z > 0 case.
@@probabilitystochasticproce2625 Yes! And to think it all started with a simple question posted on your RUclips channel. Hope all is well with you and yours.
Look at the shaded region in the graph corresponding to z < 0. For the triangular region to be integrated there, x goes from 0 --> z+y, and y goes from -z to \infinity. (Remember z is less than zero, so that -z is positive etc.)
If you do it your way, first y goes from \ -infinity to x-z . but that is for a fixed x. Then x will have to vary from \ -infinity to \+infinity outside. In other words, the 2D region to be integrated can be covered first going along x-axis, and then over y -axis(as I have it there), or the other way (as you have suggested).
![Pillai: One Function of Two Random Variables Z = sqrt [X^2 + Y^2] (Part 3 of 6)](/img/1.gif)
Only if my IIT prof was this good of a teacher
THE VIDEO IS VERY USEFUL SIR.
Good evening sir,
I m glad to watch ur lectures and learned a lot. It is a desire of mine to meet u someday . I hope we will meet someday and discuss on stochastic. Regards
Thanks. Let us hope so.
When no additional conditional was given in the first example, Z = X-Y; why didn't we consider z>0 and z
Yes, implicitly what we have there, considers both cases z>0 and z
When we consider z>0 that is possible only when x>y then why we didn't draw the region x>y and took its intersection with the region x-y
You can do what you are suggesting if you want to do z>0 and z
At T=3.58 min, the second function in the convolution should be fy(z) and not fy(y).
Probability, Stochastic Processes - 5 Minute Videos
At T=3.58 min, the second function in the convolution should be fy(z) or fy(-z)?
Merih LEBLEBİCİ.You can express
fz(z) as fx(-z)*fy(z) or as fx(z)*fy(-z). The minus sign appears only at one place.
+Merih LEBLEBİCİ
Yes, you are right. That was a slip on my part. When the random variables X and Y are independent, the pdf of Z =X-Y is given by fz(z) = fx(-z) * fy(z) .
i didn't get that z
First, draw a line for X-Y = -1. You will see that when x=0, y=1, or the line intercepts the Y-axis on the positive side as shown there. Also when y=0, x=-1 and so the line intercepts the X-axis on the negative side. This is true for any z
@@probabilitystochasticproce2625 Thanku Sir
I couldn't understand why did you chose the limits like that for z
You need to be looking at the diagram corresponding to z
This video was very helpful. I am still a little stuck on one thing: Just like in the second example, I have two random variables that are both greater than or equal to zero (X>0, Y>0). I solved for the joint density such that f(z)= integral fx(x)fy(x-z) dx. I solved the integral but I still don't know what limits to evaluate it at. Through trial and error I found that for z0, Y>0) such that the f(z)= integral fx(x)fy(x-z) dx?
+Aaron Hendrickson.
If you watch the video from T=4.16 min onward, you can see the non-negative case (X > 0, Y > 0) done separately. At T= 11.05 min, you can see the limits of integration for z > 0 and z < 0. By the way you should do the x integral first and y integral second as it is easier for z > 0 case.
Aaron. I didn't recognize you earlier!
@@probabilitystochasticproce2625 Yes! And to think it all started with a simple question posted on your RUclips channel. Hope all is well with you and yours.
Thank you so much !
em loving it .. awsum :-)
'awesome' might be a bit too much ...
Sir could u plz some more lectures for us
What do you want me to lecture on?
What does that mean ? On what topics?
why at 9:37 the limit of y is -z to infty???
Look at the shaded region in the graph corresponding to z < 0. For the triangular region to be integrated there, x goes from 0 --> z+y, and y goes from -z to \infinity. (Remember z is less than zero, so that -z is positive etc.)
why is y going from -infinity to infinity and not from x-z ?
If you do it your way, first y goes from \ -infinity to x-z . but that is for a fixed x. Then x will have to vary from \ -infinity to \+infinity outside. In other words, the 2D region to be integrated can be covered first going along x-axis, and then over y -axis(as I have it there), or the other way (as you have suggested).