What is a Random Variable?

[@ZzSlumberzZ]

1:41 I'm probably nitpicking here, but I'd like to point out that the term probability *density* is reserved for continuous distributions. The analogous term used for discrete probability distributions is probability *mass* function. This distinciton comes from the fact that conversion from discrete to continuous is not consistent mathematically (due to the extra dx term in continuous distributions.) For instance, we know that sum of the probabilities of the occurences of all possible values of random variables is 1. In the continuous case, the equivalent statement would be "area under a PDF curve is 1". The formula for the total probability *density* function has the inner product of the form f_X(x)dx. Similarly for discrete random variable, it is just P_X(x). By comparing we can conclude, for mathematical consistency, that f_X(x) must have the units of "density" rather than "mass" (the analogy probably comes from linear density = mass / length). Thought this was interesting.

[@iain_explains]

Yes, true. It was a "verbal typo". However, having said that, I personally am not a fan of the term "probability mass", because it can give the impression that the probability has a physical substantiation, that can have "mass" (which might be why I subconsciously didn't use it in the video). On the other hand, I do like the term "probability density" because it implies that "the probability" is distributed, and not necessarily evenly. ... anyway, it's just a minor point I think.

[@margaretblack5474]

I appreciate that you explicitly state that we know that P(heads) = P(tails) = 0.5 BECAUSE OF physics. I've never thought of that before, but it's really helpful to think of it in that way. There's always a reason behind even the most seemingly obvious things. Thank you!

[@iain_explains]

Glad you liked the description. Randomness can be a tricky concept when you really think about it.

[@arastooajorian9069]

I think pdf is a term for continuous random variables which shows likelihood for each point. So probability mass function is a more accurate term in discrete random variables

[@iain_explains]

Yes. Let's call it a "verbal typo". 😁

[@prasheen11man95]

for x=2 the first assumption that is 2/5 means 2 always uses so how you assumed it as 1/2

[@hariharanramamurthy9946]

so sir are saying that giving numerical analogy to probability evnet is called random variable

[@jydogeneral]

I'm not sure about the use of assumptions in the second example.. I would have thought a binomial distribution would be a good model/fit for Y?

[@JamesVestal-dz5qm]

Are the momentum, position, and energy of a molecule in a box random variables?

[@vedantjoshi1487]

Hello sir, firstly thank you for such great videos. I have a request though, can you make a video where you just put the formulas used to solve numericals so that students like me can relate the "explanation" to what is asked in exams! Thanks!

[@iain_explains]

Good suggestion. I'll try to make some videos with some worked examples.

[@vedantjoshi1487]

@@iain_explains Thank you so much sir!

[@najeebkhan2271]

Just one quick question, How can we associate probabilities with uniformly distributed random variables X= {0,1}, in a string of bits. In other words, If I wanted to send a string of bits (that are generated uniformly) over some channel, then how can I get its respective probabilities that sums to one? A little help would be appreciated.

[@iain_explains]

Sorry, I'm not really clear on what you're asking. In most communication systems the data sequence is compressed, so it can be sent efficiently. In this case, the 1s and 0s are equally likely (ie. have probability = 0.5).

[@frederickrosas5248]

Sir, can you elaborate more why P(X=2)=1/4, P(X=3)=1/2, P(X=4)=1/4? I was lost on the use of assumptions. Thank you.

[@iain_explains]

These numbers came from the assumptions. And the assumptions came from prior experience and observation. According to the particular assumptions suggested in the video, there are three types of people: those who always use WiFi when travelling on a train, those who never use WiFi when travelling on a train, and those who use it on half of the trips they make (with a random selection as to which trips they use it on). Of course this is a simplification of reality, but that is what you need to do in order to build statistical models of reasonable/practical computational size. In the video, a further assumption is suggested: that each train carriage will always contain two of the first type of people, one of the second type of people, and two of the third type of people. So, depending on what the two "third type people" have chosen to do, there will either be 2 people using WiFi (if both of the "third type" people have chosen not to use WiFi on that trip), 3 people using WiFi (if only one of the "third type" people has chosen to use WiFi on that trip), or 4 people using WiFi (if both of the "third type" people have chosen to use WiFi on that trip). Hopefully that makes it clear.

[@dmitrikazantsev3692]

Using wifi -> codeword "1", not using wifi -> codeword "0". both person1 (or 'p1'), person2(or 'p2') must use wifi always -> always "1" and "1" p3 can be "0" or "1" p4 can be "0" or "1" p5 never uses -> always "0". That gives us for p1, p2, p3, p4, p5 the following possible combinations: 11000 - 2 guys - 25% chance 11100 - 3 guys 25% 11010 - 3 guys 25% 11110 - 4 guys 25% Result: 2 guys 25%, 3 guys 25%+25% = 50%, 4 guys 25%

[@chevestong]

@@iain_explains The use of "half of the time" is what confused me. I"m still not understanding if and why this is a relevant fact/statement for this example. Using the notation by @dmitrikazantsev3692 in this comment's thread/chain, why would it matter if p3 and/or p4 only use the WiFi half of the time?

[@chevestong]

@@dmitrikazantsev3692 This truth table is what I needed to see to understand the probabilities. Thank you.

[@peterrichards5107]

Great example!

[@iain_explains]

Thanks! 😃

[@kylearigo2300]

Wait i got a question, if 2/5 people use the wifi every other day and that they're seperate, wouldnt that make "P (x=1) = 1/2"? Since there is a 50,50% that one of them would use the wifi?

[@GulzarAhmad-sw1kh]

shouldn't it be pmf instead of pdf?

[@iain_explains]

Yes, you're right. Let's call it a "verbal typo". 😁 Essentially a pdf is equivalent to a pdf, it's just that the first applies to continuous RVs and the second applies to discrete RVs. The concept is the same though.

[@yulizhang4106]

Videos help me a lot, thanks!😀

[@iain_explains]

Glad to hear that!

[@abdullahikelly9271]

great explanation I've never thought of that before whether physics has something to do with heads and tails

[@iain_explains]

Glad you liked it!

[@ayushkumarrai1117]

Sir, can you please make a video on Maximum Likelihood and Maximum A Posteriori decoding?

[@iain_explains]

Great suggestion, thanks. I've added it to my to-do list.

[@edwinr4378]

Sir ...can you please make a video on DTFT , DFT and FFT

[@iain_explains]

Do you have a specific question in mind? The FFT is just a particular efficient implementation of the DFT. They do exactly the same thing, it's just that the FFT has some of the calculations shared/parallelised in the calculation of the transform. And for the relationship with the DTFT, have you seen the following video on the channel? I discuss the relationship of the DFT/FFT to the DTFT at the 2min25sec point. kzbin.info/www/bejne/pnqpq2tqpM9smaM

[@edwinr4378]

@@iain_explains thank you sir ...I wil watch that video

[@Lisigas]

Excellent explanation, thank you!

[@iain_explains]

Glad it was helpful!

[@fratquintero]

In any random set of 5 people, the randomness of the variable (using wifi) comes from 2 users that use wifi randomly. The other 3 people has no random choices. The probability distribution function just show the probabilities of using wifi of these 2 people: P of none of them, P of 1 of them, P of all 2

[@mdayanali8885]

How P(X=2) =1/4?

[@iain_explains]

In the system, there are two people who use WiFi every day. So for X=2, it means that none of the other people are using WiFi. So, what is the probability of that happening? Well, one of the other users never uses it, and the other two use it with probability 0.5. Under the assumption that everyone decides independently, the probability that these "other two" are both _not_ using WiFi is 0.5 x 0.5 = 1/4.

[@arjunsnair4986]

Thankyou sir

[@iain_explains]

My pleasure. Glad it was helpful.

[@ziqijia5203]

Hello, sir, thank you for the explanation! But I highly suspect that you are an expert in chemistry according to your profile image. lol

[@iain_explains]

😁

[@YT-yt-yt-3]

When 2 people always use Wi-Fi, then p(x=2) should be 1 right? Why is it .25?

[@iain_explains]

No, X is the total number that are using WiFi (in this example). So some days there might be 3 using it (the two who always use it, plus one other) and then X=3 (and _not_ 2), etc.

[@YT-yt-yt-3]

@@iain_explains Thanks that helps. I assumed X is minimum number of people that are using WiFi.

[@vitumbikolowole2157]

nzeru mupenga nazo

[@DeadMyth110]

Thankyou sir

[@iain_explains]

My pleasure.