 Here is the example of the multivariate characteristics function, student moment generating function her distribution key determine nahi huti, but characteristics function her distribution key develop huti hai. So here is the example of the multivariate characteristics function, hum yaha pe bi-variate normal distribution ke saath apni example ko solve karenge. So this is the example of the multivariate characteristics function, suppose we have a bi-variate random vector, okay bi-variate random vector hum nahi yaha pe consider kiye bi-variate x and y. And we want to find its characteristics function hum nahi kya leneh iski characteristics function develop karni hai. Pehle to aap yeh dekhon ke bi-variate ka PDF kya hota? Me hi pe likhri hum bi-variate ka PDF f of xy, which is equal to 1 over 2 pi sigma of x sigma of y square root of 1 minus rho square. Pehle of the minus 1 over 2 1 minus rho square x minus mu of x divided by sigma of x whole square plus y minus mu of y divided by sigma of y whole square minus 2 rho x minus mu of x into y minus mu of y and whole divided by sigma of x sigma of y. This is the bi-variate normal distribution with the parameters of mu of x sigma of x square and mu of y sigma of y square and the rho. This is the function of the bi-variate random vector. Aap yeh hape jo hum leneh, this is the joint probability density function of x and y. Yeh aap ke paas aagye ji joint probability density function of x and y. Yeh bhi aam hape paas jointer. Baad yeh joint kiska hai? Yeh aap hape hum ne random vector ya random variable ko independent show kye hai. Jab hum independent show karte hai, it means ki unke between correlation nahi hai. Correlation kya hogi? Correlation aam hape paas 0 bhi. Yeh aap hape hum ne correlation 0 kye? Kyuki wo independent hai. Meen aap ke paas kya hogya 0? Variance aam hape paas 1 and correlation 0. Kyuki hum kya karein joint probability distribution find karein bi-variate ki? Kis ke liye independent variable ke liye? Independent meh rho 0. Variance aam hape paas sigma jo hai, wo hape paas hogi 1 and mean 0. Standard normal, independent aam hape paas. Uske according aap ke paas kya banke, yaan se aap ke paas result kya ajayega? Usko hum abhi evaluate kar lethe. Result hum aap hape paas ajayega? Factor 0 kya factor, hum aap paas 1, 1 kya sadya factor multiply ok? This is the 0, 1 square multiplied by this, to hum aap paas ajayega is the 2 pi exponential minus 1 by 2. This is 0, 1 by 2 and then x square, kyuki mu 0, variance 1, x square plus y, y square mu 0, variance 1 and rho, rho which is equals to 0 to the 3rd factor aapka poora eliminate hogya, 3rd factor hum aap paas 0 kya kal aajayega ok? Then further hum aap paas ki aage, this is the joint probability density function hai and x ki limit bi-variate normal hai, minus infinity to infinity, y ki limit minus infinity to infinity. To find the characteristics function of x and y, we need to evaluate the following integral. Tkya ab hum ne uska integral liya, janal hum aap paas kya yeh function hai. This is the psi of x t expected value of e exponential iota imaginary value t prime into x. This is the p random variable ke liye, multivariate case hai na, this is the p ab hum usko baad me bi-variate milange. So, yeh uska janal expression aap ke paas hai minus infinity to infinity upto so on minus infinity to infinity, unit into its function kya aap uski probability kao upto so on. Now here is the psi of t1, t2, kyuki bi-variate hum ne aap liya t1 and t2, isko aap open karlo for 2. Here is the exponential iota t1 x plus t2 y and the function of xy dx dy. So, function of xy, aap paas yeh hai joint of probability xy. Now here psi of t1, t2 is the characteristics function and t1, t2 are the values of the variable in the frequency domain. Aap hum ne further kya karna hai wo aap ke paas, this is the function aap uski value enter karne to solve the integral, we substitute the given joint probability function into the integral expression. Yeh aap ke paas integral hai, this is the unit into its probability or the function enter oge. Since the joint probability density function issko aap hum ne separate karna hai, separate me aap hum usko kis tari kese separate karne hai, wo hum further dekh lethe hum ne two functions uski kye, we can express at the product of two separate function. Yeh aap issko function ko separate karlo, one over pi exponential minus xk alaad, then hum ne minus y square, because this minus multiply by plus, which is equals to minus y square. Function hum aap paas aage separate karne aap usko. Now we can rewrite the characteristic integral, hum ne function ki value enter karne hai. To evaluate the integral by using the properties of the exponential function, now here use the properties of the exponential function, double integral iota t1 x1, hum ne iss funkko iss ne multiply kar liye, and second hum aare paas jo hai, wo yahan se multiply oge, oge dy dy. The integral is separate, so we can evaluate the product of two separate integrals. Aap yeh aap ke paas minus infinity to infinity hai na, yeh integral aap ke paas xk liye separate ho gaya, then next aap ke paas integral jo hai, y ke liye separate ho gaya. The first integral can be recognized as the characteristic function of the univariate standard normal distribution. This is the univariate of the standard normal distribution, yeh uska characteristics function hai. Jab hum ne usko separate kiye na, to separate karne ke baat hum aare paas, first function jo hai wo univariate standard normal distribution ka ban gaya. Aap yeh aap dek ho aap ke paas ke aura exponential of iota square t1 by 2, jo univariate me hum dek rehte iota square t1 by 2, aap yeh aap ke paas iota square hum main nazan nahi aara kyu, because the iota square which is equals to minus 1, iota square ki value hum aare paa use ho gaya, which is equals to minus 1, t1 square by 2, this is the first characteristic function, and the second integral ko hum dekhtein to uski hum, characteristics function, again iota square which is equals to minus 1, hum aare paas, second ka characteristics function aagaya. Finally, we multiply the two characteristic function together to obtain the characteristic function of the bivariate random vector, multiply aap ke paas exponential of minus, first characteristic function hum ne liye liya, then hum aare paas, second characteristic function hain usko hum ne joint ka liya, therefore the characteristics function of the bivariate random variable which is equals to this one, this is the function of the bivariate random vector. This particular example se hum ne detalmin kya kya, aur hum aare paas characteristics function develop ho gaya bivariate, standard bivariate distribution ka independently hum ne usko deal kya, aur hum kya chek ka rahe hain ke movement generating function, hard distribution ke develop nahi hota, but the characteristics function hard distribution ke develop ho gaya. Characteristics function aane ke baat further aap wo study kr logay probability distribution me, ya agar aap ne koi mathematical probabilities liye ya mathematical statistics liye, to further iski paat wo aap wahan study kr logay.