# Prob with Univariate&Bivariate&Marginal Normal Distributions

1. Mar 22, 2015

### s3a

1. The problem statement, all variables and given/known data

Problem:
In an acid-base titration, a base or acid is gradually added to the other until they have completely neutralized each other.

Let X and Y denote the milliliters of acid and base needed for equivalence, respectively.

Assume X and Y have a bivariate normal distribution with $σ_X$ = 5mL, $σ_Y$ = 2mL , $μ_X$ = 120mL , $μ_Y$ = 100mL , and ρ = 0.6.

Determine the following.:

a) Covariance of X and Y

b) Marginal probability distribution of X

c) P(X < 116)

d) Conditional probability distribution of X given that Y = 102

e) P(X < 116 | Y = 102)

Solutions given by solutions manual:
a) $ρ = cov(X,Y)/[σ_x σ_y] =0.6 cov(X,Y)= 0.6*2*5=6$

b) The marginal probability distribution of X is normal with mean $μ_x$, $σ_x$.

c) $P(X < 116) =P(X-120 < -4)=P((X_120)/5 < -0.8)=P(Z < -0.8) = 0.21$

d) The conditional probability distribution of X given Y = 102 is a bivariate normal distribution with mean and variance

μ_{X | Y = 102} = 120 - 100*0.6*(5/2) + (5/2)*0.6*102 = 123

σ^2_{X | Y = 102} = 25(1 - 0.36) = 16

e) $P(X < 116|Y=102)=P(Z < (116-123)/4)=0.040$

2. Relevant equations
$μ_{X | Y = y} = μ_x + ρ σ_X/σ_Y (y – μ_y)$

$σ^2_{Y | X = x} = σ_Y^2 (1 – ρ^2)$

Bivariate normal distribution:

$f_{XY}(x,y; σ_X, σ_Y, μ_x, μ_y, ρ) = 1/[2πσ_Xσ_Y√(1 – ρ^2)] × exp( –1/[2(1 – ρ^2)] × ((x – μ_X)/σ_X^2 – 2ρ(x – μ_X)(y - μ_Y)/[σ_Xσ_Y] + (y – μ_Y)^2/σ_Y^2) )$ for –∞ < x < ∞ and –∞ < x < ∞, with parameters σ_X > 0, σ_Y > 0, –∞ < μ_X < ∞, –∞ < μ_Y < ∞ and –1 < ρ < 1.

Univariate normal distribution:

$f(x) = 1/√(2πσ) e^{–(x – μ)^2 / [2σ^2]}$ for –∞ < x < ∞

3. The attempt at a solution
There are a few things I want to address with this post. Here they are.:

1) Would the marginal probability distribution of part b) be given by $∫^∞_∞ 1/√(2πσ_X) e^{–(x – μ_X)^2 / [2σ_X^2]} dx$ ?

2) Would this be calculable using the integral from my question “1)” above?

3) Would the conditional probability distribution of X given that Y = 102 be given by $∫^∞_∞ 1/√(2πσ_{X | Y = 102}) e^{–(x – μ_{X | Y = 102})^2 / [2σ_{X | Y = 102}^2]} dx$ ?

4) Would this be calculable using the integral from my question “3)” above?

Any input would be GREATLY appreciated!

2. Mar 22, 2015

### Ray Vickson

1) NO, obviously not. The marginal distribution of X must have x in it, but you have integrated over all x (so your final answer will not contain x anymore). You need to erase the integration sign.
2) No integration necessary for answering the question.
3) No. A conditional distribution of X must have an x in it, and you have integrated over all x (so your answer will not contain x anymore). It would be OK to just erase the integral sign.
4) Same answer as in (2).

Last edited: Mar 22, 2015
3. Mar 25, 2015

### s3a

To my knowledge, it is always the case that P(X = x) = 0 when dealing with continuous functions.

Why is the use of $f(x) = 1/√(2πσ_X) e^{–(x – μ_X)^2 / {(2σ_X^2)}} (for –∞ < x < ∞)$ for part b) and $f(x) = 1/√(2πσ_{X | Y = 102}) e^{–(x – μ_{X | Y = 102})^2 / {(2σ_{X | Y = 102}^2)}} (for –∞ < x < ∞)$ for part d) not contradicting my above statement?

Also, the sheet of paper that my school gives, which has the table for the normal distribution's Z values and probabilities, states that $Φ(z) = P(Z ≤ z) = ∫ _{-∞}^x 1/√{2π} e^{-1/2 × u^2} du$. How does this relate to the two f(x) functions I mentioned in the above paragraph (=second bunch of text that doesn't skip lines in this post)?

P.S.
Sorry for the formatting screw-up of my initial post.

4. Mar 26, 2015

### Ray Vickson

You need not worry about any contradictions regarding $f_{X|Y}(x|Y=y)$, even though $P(Y=y) = 0$ for every $y$. You can think of $f_{X|Y}(x|Y=y)$ as a limit of
$$f_{X|Y}(x|y < Y < y + \Delta y) = \frac{f_{XY}(x, y < Y < y+\Delta y)}{P(y < Y < y+\Delta y)}$$
as $\Delta y \to 0$.

I don't understand the rest of your question. Surely you know how to covert a probability such as $P(X_{N(\mu, \sigma)} \leq x)$ into the standard $P(Z \leq z)$, where $Z \sim N(0,1)$. You must have used it dozens of times already in your studies.

5. Mar 26, 2015

### s3a

Unless I'm misunderstanding something right now, it seems we're miscommunicating.

I feel that I have some fundamental terminology confusions.

Basically,
1. Is f(x) called a (continuous) probability distribution function?

2. In the continuous case, is a probability distribution function the integral from -∞ to x of a probability density function (such that the probability distribution function is a cumulative distribution function)?

3. Is it the case that f(x) = P(X = x), or is it the case that [ii] f(x) = P(X ≤ x) = P(X < x)? For what it's worth, I'm thinking that [ii] is correct.

4. Let's say I looked at a table of values for a normal distribution function, and I took some value of Z, z, and then used that to find X by using Z = (X - μ)/σ to get X = x = z * σ + μ, would computing f(x) = f(z * σ + μ) give me P(X ≤ x) = P(X < x)?

5. So, finding the marginal distribution function of X just means I should ignore the all the Y-related stuff like σ_Y, whereas a conditional probability distribution of X is similar in the sense that it also requires that I use the univariate normal probability distribution function instead of the bivariate one except that I take the Y-related stuff into account by finding a single μ and σ given by the $μ_{X|Y=y} = μ_X + ρ σ_X/σ_Y(y–μ_Y)$ and $σ^2_{Y|X=x} = σ^2_Y(1–ρ^2)$ formulas, right?

6. Mar 27, 2015

### haruspex

AFAIK, "probability distribution function" is not a defined thing. You can have a probability density function (PDF), and a corresponding cumulative distribution function (CDF). In the case that the first exists, it is the derivative of the second.
f(x0 (lower case) is usually reserved for the density function, and F(x) for the CDF. F(x) = P(X<x).
Not ignore, exactly. It means you need to integrate over y (not over x).

7. Mar 27, 2015

### s3a

Would the full answers to parts b and d involve an integral? (I say "full answers", because the answers given seem like sloppy shortcuts for those who already have a good understanding of the material.)

8. Mar 27, 2015

### Ray Vickson

For a continuous random variable $X$, $P(X=x) = 0$ for all $x$. However, there is a probability density $f(x)$ such that for small $\Delta x > 0$ we have
$$P(x < X < x + \Delta x) = f(x) \, \Delta x + O((\Delta x)^2)$$,
so $P(x < X < x | \Delta x) \doteq f(x) \Delta x$, accurate to first order in the small quantity $\Delta x$. The cumulative distribution function (nowadays sometimes called just the distribution function---erasing the word "cumulative") is $F(x) = P( X \leq x)$. (However, some writers use $X < x$ instead; the difference is immaterial for purely continuous $X$, but the distinction is important when the random variable is discrete, or mixed, partly continuous and partly discrete.)

(2): YES, for continuous $X$

(3): Everything you write there is false for a continuous random variable (that is, if you have already used the symbol $f(x)$ to denote the density). It is customary to use a lower-case letter for the density and the corresponding upper case letter for the (cumulative) distribution. This is convention, not law, so you are allowed to violate it provided that you explain your notation first.

(4) No. If $f(x)$ means the density, then $P(X \leq x) = \int_{-\infty}^x f(t) \, dt$ is certainly not equal to $f(x)$ or anything like it. If $\phi (z)$ denotes density of the standard normal random variable $Z \sim N(0,1)$---that is, $\phi(z) = \exp (-z^2/2)/\sqrt{2 \pi}$ --- then for $X \sim N(\mu,\sigma)$ we have $X = \mu + \sigma Z$, $f(x) = \frac{1}{\sigma} \phi((x - \mu)/\sigma)$ and $F(x) = \Phi((x-\mu)/\sigma)$, where $\Phi(z) = P(Z \leq z)$. The function $\Phi$ is tabulated and is also available in many hand-held calculators and in spreadsheets, etc.

(5) Sort of right. The marginal distribution of $X$ integrates over all the other random variables that accompany $X$ in a multivariate distribution. In the case of a multivariate normal with parameters $\mu_x, \mu_y, \sigma_x^2, \sigma_y^2, \sigma_{xy}$, the integration can be done explicitly, and when you do that you end up with the marginal density
$$f_X(x) = \frac{1}{\sigma \sqrt{2 \pi}} \exp\left( - \frac{(x - \mu_x)^2}{2 \sigma_x^2} \right)$$
One way to say it is that you ignore $\mu_y, \sigma_y^2 , \sigma_{xy}$---anything with a "y" in it. However, the fact that you can ignore those things is a provably correct result, not just intjuition.

The case of $f(x|y)$ is different: you must have some of the y-related quantities in the formula, and the correct expressions are the ones you wrote. It is important to remember that in this case "y" acts like a constant parameter in the distribution of $X$, while $x$ can vary over all $\mathbb{R}$.

9. Mar 9, 2016

### punsb

Bro, where did you find the solution manual for this 6th edition. Or whichever edition you have. Please please share.