[Music] and hello and welcome to today's lecture in last class we had uh discussed the normal distributions towards the end of last lecture so we'll Begin by briefly recapping what we had discussed so for a normal distribution your probity density function FX Has the expression 1 by Sigma < tk2 piun exponential x - mu 2 by 2 Sigma s where X can have any values between minus and plus infinity okay so we had derived that for so as you see that there are two parameters in this normal distribution poity density function This Is mu and sigma and we had derived that for a r normal random variable expectation of X will turn out to be mu and the standard or variance of X will come out to be Sigma Square okay also for a normal distribution okay because it is symmetric okay your mean median and mode give you the same value okay are equal okay now from this random variable you can convert it into a standard normal variable which is typically given by this expression Z is equal to x - mu by Sigma if you do this then you get F instead of f of x you will have F of Zed which turns out to be 1 by < tk2 piun E the^ minus z s by 2 okay okay and in this case the normal distribution so this the regular normal distribution has parameters mu mean mu so this is your value mu and Varan Sigma Square okay so roughly 3x so 3 3 Sigma will mu +- 3 Sigma will contain all your data okay for this for the standard normal variable so this is written as the standard normal variable so this has mean of zero and variance of one okay so this will be centered at the origin okay and it will be symmetric okay so the cumulative distribution function the cumulative distribution function is given by 5 of X okay which is given by 1 by < tk2 piun minus infinity tox e^ - y^ 2 by 2 Dy okay so if you have any X so probability of let's say x which is great less than b can be converted to a standard normal variable is equal to to F of B minus mu by Sigma okay so if x has mean of mu and variance of Sigma squ then this P of X less than b is nothing but F of B minus mu by Sigma similarly probability of a less than x less than b will give you f of b - mu by Sigma minus 5 of a - mu by Sigma one more thing because your distribution is symmetric okay so let's say this is X and this is min - x this is your origin okay so five of Min - x which is this area Okay will be equal to this okay so you can write 5 of minus X as nothing that 1 - 5 of X you can see it visualize it from this plot 5 of - x is this area 5 of X is this entire area and 1 is the entire area under the curve so 1 minus 5 of X will give you this area which is equal to this area so this is another useful expression to remember okay now there's another interesting idea okay so if you have random variables X1 X2 X3 dot dot dot xn okay we which are all normal random variables okay with means of mu1 mu2 mu n and sigma 1 s and variances Sigma 1 square Sigma 2 square dot do Sigma n Square then for let's say if you have a variable X which you define as summation of x i okay so the moment generating function for this x will be F of T is exponential of e ^ TX now let's assume also that these random variables are independent okay so your moment generating functions becomes 5 of tal to e^ TX in last class we had derived that the moment generating function is gives you the expression exponential of Mt + Sigma s t² by 2 okay so this is for single variable okay so for this case where X is summation x i I can write as e to the power e ^ T summation of XI and E to the^ T summation of XI can be written as e to the^ expectation of Pi of e ^ TXI okay where Pi is Pi of XI I equal to 1 to n so Pi is nothing but the product that is so this is e^ tx1 e^ tx2 dot dot dot okay so I can take this product out and I can write down as Pi of expectation of e ^ TXI okay and for this I know that F of T is has gives you this expression okay so this is nothing but Pi of okay e to the power mu i t + Sigma i² t² by 2 okay and you have a product okay So eventually you will get this 5 of t as e to the^ summation of mu t + Sigma i s t² by 2 okay this tells you this tells you that for the random variable which is defined as the sum of these independent random variables which are normal distributions okay normal random variables the mean so X will follow a normal distribution with mean mu okay so X will follow normal distribution with mean summation of mui and variance summation of Sigma I squ okay so let us consider as an example where this concept might be very useful okay so imagine you have rainfall okay the yearly rainfall okay yearly rainfall okay in a city is a normal random variable with mu equal to 12 CM and sigma equal to 3 cm okay and you asked the question okay what is the probability okay of rainfall for the next 2 years exceeding 25 cm okay what you see here is that since the rainfall in a given year is a normal random variable and there is no correlation between the rainfall in a given year and the rainfall in the next year so you can assume that the rainfall in each year is a standard normal variable and these variables are independent of each other okay so I can write X1 is the normal random variable of rainfall in year one and EXT 2 is similarly the normal RV of rainfall in year 2 okay so what we have been asked to calculate is the probability of X1 + X2 greater than 25 is what we have been asked to calculate so this is this is what we have been asked to calculate okay so now invoking the proof the theorem that we just proved that for a variable X which is defined as summation x i okay okay your mu where XI are normal random variables your mu is sumission mui and sigma square is sumission Sigma I sare okay so if we use this particular idea then what we should have is the variable X1 + X2 should have mean = 12 + 12 that is equal to 24 and variance okay equal to 3² + + 3² = 18 okay so we want to calculate probability of X1 + X2 greater than 25 okay so I can convert this into a standard normal variable so let's say this is X so I can Define X is = to x - mu by Sigma which is equal to x - mu is 24 because this random variable has a mean of 24 and and sigma of so Sigma square is 18 okay it is root of 18 is the sigma okay so what we can then do okay so probability of X1 + X2 greater than 25 is nothing but probability of Z greater than 25 - 24 by otk of 80 okay so this is you know roughly if you look up the tables the normal distribution tables you will see this probability come out to be4 okay we can ask another question okay let us take another question okay where what is the probability that okay that this years's rainfall exceeds that of next year okay by more than 3 cm okay so again if X1 is for the normal random variable for year 1 and X2 is for year two then what we have been asked to determine is probability of X1 - X2 greater than 3 okay so once again we can use the previous uh you know Theory and say that okay so the random variable X = X1 - X2 will have a mean of 12 - 12 which is equal to 0 okay and a variance of 3 s + 3² = 18 okay so just thus probability of X1 - X2 greater than 3 is nothing but okay probability of X1 - X2 - 0 is the mean by otk of 18 greater then 3 byun of 18 okay is probability of Zed greater than 3x otk of 18 and this probability if you look up the tables it comes out to be approximately 25% okay so this is how you can make use of random variables of multiple random variables which are independent of each other and find out the probability associated with their sum or and or sub or or subtraction of the two random variables so on and so forth okay so that brings our discussion of Random variables or normal random variables to a close I would like to discuss one more random variable which is called an exponential random variable okay for an exponential random variable variable so your FX is defined as Lambda e^ minus Lambda X when X is great greater equal to Z and equal to Z otherwise okay so I can see whether so summation of FX would mean integral from 0 to Infinity Lambda e^ minus Lambda X by DX okay I can take Lambda out and it isus Lambda X by minus Lambda 0 and infinity which is equal to one okay so this fulfill summation of probability is all probity this is one okay I can determine the cumulative distribution function okay f of x is nothing but okay is probability X less equal to X is equal to 0 to X Lambda minus Lambda X DX which turns out to be okay Lambda 1 minus E the^ minus Lambda X okay so your cumulative distribution functions is this which means that probability of X greater than x will be equal to e ^ minus Lambda X so we'll make use of this probability slightly later okay but for this exponential random variable we can calculate the moment generating function okay so the moment generating function will be exponential e ^ TX is equal to 0 to Infinity e^ TX okay into Lambda e^ minus Lambda X DX okay equal to I can take Lambda out 0 to Infinity the power minus Lambda minus t into X DX okay equal to so this should give you a value of okay e^ minus Lambda minus TX by Lambda minus t with a minus sign and this is from 0 to Infinity so this should give you a value of Lambda by Lambda minus t Okay so 5 of T is this which would mean that 5 of 0 okay okay t equal 0 is equal to 1 5 Prime of t Okay 5 Prime of T is okay so this is Lambda minus 1 is equal [Music] to 1 squ okay equal to minus Lambda by Lambda - t s okay so if I write it as 5 Prime of T okay let's see 5 0 is equal to 1 okay this can be written okay so you can write minus Lambda by T minus Lambda okay T minus minus okay so this can be written as 5 Prime of T can be written as minus Lambda into DDT of tus Lambda ^ -1 is = to- or- + Lambda by T - Lambda s = Lambda by Lambda - t² okay so 5 Prime of 0 which is nothing but exponential of expectation of x X returns your value of Lambda by Lambda Square = to 1 by Lambda okay so expectation okay expectation of X gives you a value of 1 by Lambda okay we can accordingly do five double Prime of T So 5 of T is gives you a value of Lambda by Lambda minus t Okay F Prime of T is Lambda by Lambda - t² okay 5 Prime of T will give you okay 2 - 2 1 to Lambda - t cubed is equal to 2 Lambda by Lambda - t cubed okay so 5 Prime of 0 will be 2 Lambda by Lambda cubed is equal to 2 by Lambda s so variance of of X will give you e of so this is e of x² okay e of x² minus E x² = 2 by Lambda s - 1 by Lambda s = 1 by Lambda sare okay so variance of X gives you 1 by Lambda squ and expectation of X okay so this gives you expectation of x = 5 Prime of 0 is = to 1 by Lambda okay so one very important property one very important property of of uh exponential distributions this is slide number nine so one very important property of exponential distributions is its exponential distributions okay has this property that this distribution is memoryless in other words so let us take a case where we considering the lifetime of a battery of a car or lifetime of a disc or so on and so forth so we would typically expect that if we know that the battery was operating okay as of today it will probably depend so how long much longer how much longer it will last will depend on how much it was operating now however in the case of a memoryless distribution so you can write probability of X is the lifetime greater than t + S given X is greater than T in other words it has lasted at least for T amount of time what is the probability it will last for another s amount of time so this one would ideally expect to depend on this T but this has the probability this is does not depend on T and this is simply probability of X greater than x so this is the memory list so this is true for T Comm s greater equal to Z okay so this is the memory lless property of of an exponential distribution okay and we will see why this is so okay so we can see we have calculated we know that FX okay is given by 1 - e^ minus Lambda X okay which would mean that probability of X greater than x is equal to e ^ minus Lambda X okay so if probability so which means that probability of okay X so greater equal to X is e to l so probability of X greater than S + T would simply be equal to e^ minus Lambda of s + T okay and this I can break it down e^ minus Lambda s into e^ minus Lambda T is equal to probability of X greater than S into probability of X greater than T okay now if this is so then from this and this I can write down so probability of X greater than X+ T given X greater than T is simp simply probability of X great than x + T okay so you can simply write down this divided by this is so probability of X greater than t + S given X greater than T is equal to probability of X greater than S + t s + T by probability of X greater than T and this gives me the value probability of X greater than S okay so this proves that this is the memoryless property of this particular distribution okay and let us take one particular example to discuss discuss how it is useful okay let's say the lifetime okay of a battery a car battery okay is exponentially distributed with a average of 10,000 kilm okay if a person wishes to go on a trip of 5,000 km what is the chance that she will be able to do it without having to change the battery okay okay so what you're given is the average lifetime so average lifetime okay is equal to okay so 10 K kilometers okay so let's say I'm if I if I put the average lifetime in in units of thousands of kilomet thousands of kilometers okay so my lifetime is 10 okay and I know the expectation so expectation of this exponential distribution is 1 by Lambda and this is same as this 10 which would mean my Lambda is 1 by 10 okay so I want to calculate probability of Lifetime greater than 5 okay which is nothing but 1 minus okay F of 5 = to e to the power so we know know e the^ Lambda into 5 okay which is nothing but e ^ minus Lambda * 5 equal to e ^ minus half okay so this is what you see that this distribution does not depend as to how long the battery lasted before but it only depends on what is the average and using that you can calculate the lifetime okay with that we conclude our discussion of quity distributions so across multiple lectures we discussed about binomial and special random variables including the binomial random variable the poison random variable the uniform random variable the normal distribution and the exponential distribution so from next lecture onwards we will start discussing about sampling distributions thank you for your attention and I I B