 Dear students, let me give you an example as to how you would determine the probability of an event in the case of a continuous random vector. Suppose that we have the joint PDF, small f of x1, x2 equal to six x1 square into x2 where x1 goes from zero to one and x2 also goes from zero to one and f of x1, x2 is equal to zero elsewhere. If this is the PDF of the random vector capital x1, capital x2, where both of them are continuous random variables, then suppose that we are interested in computing the probability that capital x1 lies between zero and three by four and capital x2 lies between one by three and two. All right, this is our question. Now let's see how we deal with it. The first thing is that we will deal with this later. First of all, don't you want to check that the function given to us is really a PDF? If you want to do that, then how will you do it? Obviously, you should integrate six x1 square into x2 with respect to x1 and x2 and you should take the entire limit for each one of them as the limits of those integrals. You can check it yourself. Assuming that it is a PDF, now let us try to find this particular probability that we are wanting to find. All right, the required probability will be the integral from one by three to two. I have applied the limits for x2 and the next integral that is inside the first one because what we just said is that dx2 is attached to the end of the expression but generally we do not put the bracket, we just write it like this, double integral. Now what is the limit we have to apply for the second one? The limits of x1. So that is going from zero to three by four. We want to see what is the probability of this happening. All right, now let us continue. This expression has come, so it is obvious that f of x1, x2, which is written in it, we will insert the expression of six x1 square into x2. Note that the joint PDF that was given to you, that was not going from zero to two, i.e. x1 was not going from zero to two, x2 was not going from zero to two. But what we have been told is that we should take out this probability, we are saying that x2 is between one by three and two, how do we deal with this one? We split the first integral. As you can now see on the screen, the required probability is the integral from one by three to one as far as the variable x2 is concerned plus the integral from one to two as far as the variable x2 is concerned. The rest was not the same as before. In fact, you are seeing that the second expression is written in it, after the double integral, zero is written. And it should be written, and it will be written. Because as I told you earlier, how we have to express it and how it has been done at this time, it is f of x1, x2 is equal to zero elsewhere. Zero to one, zero to one, both the variables, zero to one, zero to one in the area, so your expression has six x1 square x2. But if there is any other area of the x1, x2 plane, this function is equal to zero there. So after splitting the integral, if our second integral is going from one to two, then f of x1, x2 is equal to zero. All right, now we can try to solve these integrals. If we first look at the first one, one by three to one, zero to three by four, six x1 square x2, and then dx1 dx2. So now it is very important that we keep this in mind that first we solve the inner one, and after that the outer integral will be solved. So inside that is with respect to x1. So if we look at x1, then six x2 that is acting as a constant. Because x1 is not in it, six is a constant and x2 acts as a constant. So the inner integral, six x2 will come out of that. But it will not come out, six will come out of all of it, but x2 will not come out of the first integral, which is going from one by three to one, because that is with respect to x2. So x2 cannot come out of that, but our inner integral came out of that. I am talking about the first integral now. The second integral is written as zero. Now when you solve the inner integral of this zero, what is the integral of zero? Zero ka integral kya hai? It is c, where c is a constant. Am I right? Of course it is correct. Do you not know that the derivative of a constant is equal to zero? Agar constant ka derivative zero hai, so zero ka integral c hai. And after that, when you apply the upper and lower limit, students do not apply the limit on the constant, that is, the zero to three by four written there, so you say that c, you apply the upper limit first, then you write three by four, and then the lower limit is minus zero. So the answer is three by four. This is not correct. C agar hai saath koi x1 attached hi nahi hai, to aap likhenge c minus c and it is equal to zero. So therefore, now we have the first integral as it is, one to two, aur ye jo beech wala tha uspuri tiz ki jaga pe zero, aur uske saath dx2. But please note kya ab yeh saari guftabu, jo mai ne x1 ke liye ki, that will be repeated for x2. To wo bhi zero hi hojayaega, and therefore that entire double integral of zero with those limits is zero. Coming back to the first one, six to bahar aagya tha, aur uske baad aap ka jo pehla integral hai, that is going from one by three to one, aur uske baad aap ne x2 likhawa hai, kyu ke wo uske aage jo integral hai usmise bahar aagya wa tha, aur uske baad you have the integral from zero to three by four of x1 square, and you are integrating with respect to x1. To ab kya hota hai, what is the integral of x1 square? Obviously it will be x1 raised to three divided by three, aur uske baad aap limit-supply kar diye, solve kar liye, and I think I have said enough. Iske baad just keep on solving the way you should solve integrals, and doing this you will ultimately, after carrying out all the steps one by one by one, ultimately you obtain the probability of this particular event as being equal to three by eight. So this is the method by which you may compute the probability of any event for that situation where you have a continuous random vector and you are wanting to do it. Sara jo kuch main aapke saamne rakha that was with reference to two random variables only. Agar aap ke paas doki bhajai teen hota, x1, x2, x3, aur ishi tara ki cheezen hoti, to kya kar na chahiye tha? I am sure you must have got it. Ek aklota variable, continuous variable, ek integral, two continuous variables, double integral, three continuous variables, triple integral. So this is the mechanism. Coming back to this bivariate situation, I would like you to note ke aapne abhi e bhi e jo probability nikali hai, the answer of which has come out to be three by eight. Zahire ke three by eight ko aap decimal me bhi likh sakte hai, percentage form me bhi likh sakte hai, but what is the geometric interpretation? It is the volume under the surface defined by the function f of x1, x2 equal to 6x1 squared into x2. It is under this surface above the rectangular area of the x1, x2 floor defined by x1 going from zero to three by four and x2 going from one by three to one. Note ke me ne one by three to two nahi kaha, yu ke me pehle batah chuki hu ke one se two ke beech me to f of x1, x2 is equal to zero. Abhi e jo abhi kaha isko phir dubara sunn liye je we are having two variables so we are dealing with the just one x-axis and the probability curve being along the y-axis meaning that the ordinates define that curve is what we have a floor and the ordinates are giving us the surface. The total volume under which we will be one but the volume of our event that event that we are interested in is equal to three by eight.