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I have a question from my stats class that I am confused about how to proceed with. I have a general idea of what I am to do but I am not sure how to start.

The question is about a car that is moving on a road. It's beginning position is b, and b is normally distributed ~N(0,6). At b the car drives around, the distance the car drives is d with d~N(0,5). If d is positive, the car went up. Otherwise it moves to the down. Given that b and d are statistically independent and the car's final location is 6 or c = 6 (c being the final placement of the car). Find the most likely location b or where the car first began moving.

From what I understand, c=6 is just an observation and we are to find P(b|d=6) with b ~N(0,6) and d~(0,5), maximizing for the probability that d = 6. But I'm not sure how to proceed from that.

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EDIT

I messed up the notation of the normal distribution. All occurrences of the variance $5^2$ and $6^2$ must be replaced by $5$ and $6$ respectively. This also affects the calculations

I'll try and fix it tomorrow since I don't have access to octave right now.


I'll give it a try, but I am not 100% sure. I think what you are looking for is actually the maximum of $P(B|C=6)$.

You can start by noting that

$P(C|B,D) = \delta(C-(B+D)) $,

which is just a formal expression for "C is the sum of B and D". Also we can state that $P(C)$ is normally distributed like this: (see here):

$P(C) \sim \mathcal{N}(0,5^2+6^2)$

Then by marginalization we also have the conditional probability of C conditioned on B.

$$\begin{eqnarray} P(C|B) &=& \int P(C|B,D)\cdot P(D) = \int \delta(C-(B+D)) P(D) dD \\ &=& \int \delta(D-(B-C))\cdot \frac{1}{\sqrt{2\pi\cdot 5^2}} \exp\left(\frac{-D^2}{(2\cdot 5^2)}\right) dD \\ &=&P(D=C-B)\\ &=&\frac{1}{\sqrt{2\pi\cdot 5^2}} \exp\left(\frac{-(C-B)^2}{(2\cdot 5^2)}\right) \end{eqnarray}$$ which is normally distruted as per problem formulation.

Then using Bayes theorem you obtain $P(B|C)$ from the above:

$P(B|C) = \frac{P(C|B)\cdot P(B)}{P(C)}$

In the problem P(B) is given as a normal distribution $P(B)= \frac{1}{\sqrt{2\pi\cdot 6^2}} \exp\left(\frac{-B^2}{(2\cdot 6^2)}\right)$.

for the case $P(B|C=6)$ the C terms evaluate to a constant, meaning you can just plug in $C=6$. Also $P(C)$ is a constant given by the normal distribution $\mathcal{N}(0,5^2+6^2)$ as mentioned above, but that does not matter since you are only interested in the most likely value of $B$. That leaves you with an expression of $B$. Here you correctly said to evaluate the maximum. So let us have a look.

Leaving out the constants of proportionality we are left with an expression of $B$ like this:

$P(B|C=6) \propto \exp\left(\frac{-(6-B)^2}{(2\cdot 5^2)}\right) \cdot \exp\left(\frac{-B^2}{(2\cdot 6^2)}\right)$

Plotting it out with Matlab / OCTAVE using this code:

B = linspace(-20,20,1024);
Prob = exp( -(6-B).^2 / (2*5^2) ).* exp( -(B).^2 / (2*6^2) );
[maxval, maxPos] = max(Prob);
Bmax = B(maxPos)
#plotting
figure
plot(B,Prob);
title("Posterior(unnormalized)");
xlabel("Value of B");

gives me this graph:

enter image description here

showing a maximum at $B\approx 3.54$.

Again: not sure if everything is 100% correct, but I hope it helps.

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  • $\begingroup$ I just want to say thank you for your input and with your help I've realized what the problem was asking. I think what the question is asking is deriving the Maximum a posteriori of the a univariate normal (the summed normal C). I'm not 100% sure either on my answer. Thank you for your help and taking the extra step with the code. $\endgroup$ – James S. Nov 5 '18 at 5:17
  • $\begingroup$ @JamesS. Maybe yes. But the way the question is phrased it asks for the most likely starting position given the end position. This is what I tried to calculate from the $P(B|C)$ $\endgroup$ – geo Nov 5 '18 at 6:28

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