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For a standard Brownian motion (Bt:t≥0)\left(B_{t}: t \geq 0\right) a) Find $\mathbb{...
Apr 30, 2024
For a standard Brownian motion (Bt:t≥0)\left(B_{t}: t \geq 0\right) a) Find \mathbb{P}\left(B_{2}-B_{1}>1 \mid B_{0.5} \leq 2\right).
Solution by Steps
step 1
Use the Markov property of Brownian motion, which states that given the present, the future is independent of the past
step 2
Calculate the conditional probability \mathbb{P}\left(B_{2}-B_{1}>1 \mid B_{0.5} \leq 2\right) using the fact that B2−B1B_{2}-B_{1} is independent of B0.5B_{0.5}
step 3
Recognize that B2−B1B_{2}-B_{1} is normally distributed with mean 0 and variance 2−1=12-1=1 since increments of Brownian motion are normally distributed with mean 0 and variance equal to the time increment
step 4
Calculate the probability using the standard normal distribution: \mathbb{P}(Z > 1) where ZZ is a standard normal random variable
step 5
Use standard normal distribution tables or software to find \mathbb{P}(Z > 1)
Answer
\mathbb{P}\left(B_{2}-B_{1}>1 \mid B_{0.5} \leq 2\right) = \mathbb{P}(Z > 1)
Key Concept
Independence of increments in Brownian motion and the properties of the standard normal distribution.
Explanation
The probability that B_{2}-B_{1}>1 given B0.5≤2B_{0.5} \leq 2 is the same as the probability that a standard normal variable is greater than 1, due to the independence of increments in Brownian motion.
P(Z>1)
Solution by Steps
step 1
Identify the distribution of Z Z : Since Z Z is not explicitly defined in the context, we assume Z Z is a standard normal random variable, which means Z∼N(0,1) Z \sim N(0,1)
step 2
Calculate the probability \mathbb{P}(Z > 1) : This is the area under the standard normal curve to the right of z=1 z = 1
step 3
Use standard normal distribution tables or software to find the area: The tables or software will provide the area to the left of z=1 z = 1 , which we denote as Φ(1) \Phi(1)
step 4
Subtract the area to the left from 1 to find the area to the right: \mathbb{P}(Z > 1) = 1 - \Phi(1)
Answer
\mathbb{P}(Z > 1) = 1 - \Phi(1)
Key Concept
Calculating the tail probability of a standard normal distribution
Explanation
The probability \mathbb{P}(Z > 1) represents the area under the standard normal curve to the right of z=1 z = 1 . This is found by subtracting the cumulative distribution function value at z=1 z = 1 from 1.
1−Φ(1)
Solution by Steps
step 1
To calculate \mathbb{P}(Z > 1) , we need to understand that Z Z is a standard normal random variable with mean 0 and standard deviation 1
step 2
The cumulative distribution function (CDF) of a standard normal random variable at a point z z is denoted by Φ(z) \Phi(z) , which gives the probability that Z Z is less than or equal to z z
step 3
To find \mathbb{P}(Z > 1) , we subtract the CDF at z=1 z = 1 from 1: \mathbb{P}(Z > 1) = 1 - \Phi(1)
Answer
\mathbb{P}(Z > 1) = 1 - \Phi(1)
Key Concept
Standard Normal Distribution and its CDF
Explanation
The probability of a standard normal variable being greater than a certain value is found by subtracting the CDF of that value from 1.
◊What is the value of the standard normal distribution function at z=1? 1−Φ(1)1 - \Phi(1)⍭ Generate me a similar question◊
Consider the process Xt=Bt−tB10≤t≤1. X_{t}=B_{t}-t B_{1} \quad 0 \leq t \leq 1 . i) Derive E[Xt]\mathbb{E}\left[X_{t}\right] and Cov⁡(Xt,Xs)\operatorname{Cov}\left(X_{t}, X_{s}\right) for some fixed t,s∈[0,1]t, s \in[0,1] and specify the joint distribution of XtX_{t} and XsX_{s}. ii) Calculate \mathbb{P}\left(X_{0.15}>0.2\right). iii) Find E[Xs∣Xt]\mathbb{E}\left[X_{s} \mid X_{t}\right] for some fixed s,t∈[0,1]s, t \in[0,1] such that 0<s<t, and use this to give meaning to the expectation of XsX_{s} given the event that Xt=1X_{t}=1.
Solution by Steps
step 1
To derive E[Xt]\mathbb{E}[X_{t}], we use the linearity of expectation and the properties of Brownian motion
step 2
Since Xt=Bt−tB1X_{t} = B_{t} - tB_{1}, we have E[Xt]=E[Bt]−tE[B1]\mathbb{E}[X_{t}] = \mathbb{E}[B_{t}] - t\mathbb{E}[B_{1}]
step 3
Knowing that E[Bt]=0\mathbb{E}[B_{t}] = 0 for any tt, we get E[Xt]=0−t⋅0=0\mathbb{E}[X_{t}] = 0 - t \cdot 0 = 0
step 4
To find Cov⁡(Xt,Xs)\operatorname{Cov}(X_{t}, X_{s}), we use the definition of covariance and properties of Brownian motion
step 5
We have Cov⁡(Xt,Xs)=E[(Xt−E[Xt])(Xs−E[Xs])]\operatorname{Cov}(X_{t}, X_{s}) = \mathbb{E}[(X_{t} - \mathbb{E}[X_{t}])(X_{s} - \mathbb{E}[X_{s}])]
step 6
Substituting XtX_{t} and XsX_{s} and expanding, we get Cov⁡(Xt,Xs)=E[(Bt−tB1)(Bs−sB1)]\operatorname{Cov}(X_{t}, X_{s}) = \mathbb{E}[(B_{t} - tB_{1})(B_{s} - sB_{1})]
step 7
Using the properties of Brownian motion, we find Cov⁡(Xt,Xs)=min⁡(t,s)−ts\operatorname{Cov}(X_{t}, X_{s}) = \min(t, s) - ts
step 8
The joint distribution of XtX_{t} and XsX_{s} is bivariate normal with mean vector [0,0][0, 0] and covariance matrix [t−t2amp;min⁡(t,s)−tsmin⁡(t,s)−tsamp;s−s2]\begin{bmatrix} t - t^2 &amp; \min(t, s) - ts \\ \min(t, s) - ts &amp; s - s^2 \end{bmatrix}
Answer
E[Xt]=0\mathbb{E}[X_{t}] = 0, Cov⁡(Xt,Xs)=min⁡(t,s)−ts\operatorname{Cov}(X_{t}, X_{s}) = \min(t, s) - ts, and the joint distribution of XtX_{t} and XsX_{s} is bivariate normal with the specified mean vector and covariance matrix.
Key Concept
Expectation and covariance of transformed Brownian motion
Explanation
The expectation of XtX_{t} is zero due to the properties of Brownian motion, and the covariance is derived from the properties of Brownian motion and the definition of covariance. The joint distribution is bivariate normal due to the linear transformation of Brownian motion.
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step 1
To calculate \mathbb{P}(X_{0.15} > 0.2), we first find the distribution of X0.15X_{0.15}
step 2
Since Xt=Bt−tB1X_{t} = B_{t} - tB_{1}, we know that X0.15X_{0.15} is normally distributed with mean 0 and variance 0.15−(0.15)20.15 - (0.15)^2
step 3
We standardize X0.15X_{0.15} to find the probability: Z=X0.15−00.15−(0.15)2Z = \frac{X_{0.15} - 0}{\sqrt{0.15 - (0.15)^2}}
step 4
We calculate \mathbb{P}(Z > \frac{0.2}{\sqrt{0.15 - (0.15)^2}})
step 5
Using standard normal tables or a calculator, we find the probability corresponding to the standardized value
Answer
\mathbb{P}(X_{0.15} > 0.2) is the probability corresponding to the standardized value calculated in step 4.
Key Concept
Standardizing a normal random variable
Explanation
To find the probability that a normal random variable is greater than a certain value, we standardize it and use the standard normal distribution to find the probability.
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step 1
To find E[Xs∣Xt]\mathbb{E}[X_{s} | X_{t}] for 0 < s < t, we use the properties of conditional expectation in the Gaussian case
step 2
We have E[Xs∣Xt]=E[Xs]+Cov⁡(Xs,Xt)Var⁡(Xt)(Xt−E[Xt])\mathbb{E}[X_{s} | X_{t}] = \mathbb{E}[X_{s}] + \frac{\operatorname{Cov}(X_{s}, X_{t})}{\operatorname{Var}(X_{t})}(X_{t} - \mathbb{E}[X_{t}])
step 3
Substituting the known values, we get E[Xs∣Xt]=0+s−stt−t2(Xt−0)\mathbb{E}[X_{s} | X_{t}] = 0 + \frac{s - st}{t - t^2}(X_{t} - 0)
step 4
Simplifying, we find E[Xs∣Xt]=s−stt−t2Xt\mathbb{E}[X_{s} | X_{t}] = \frac{s - st}{t - t^2}X_{t}
step 5
To give meaning to the expectation of XsX_{s} given the event that Xt=1X_{t} = 1, we substitute Xt=1X_{t} = 1 into the conditional expectation formula
step 6
We get E[Xs∣Xt=1]=s−stt−t2⋅1\mathbb{E}[X_{s} | X_{t} = 1] = \frac{s - st}{t - t^2} \cdot 1
Answer
E[Xs∣Xt]=s−stt−t2Xt\mathbb{E}[X_{s} | X_{t}] = \frac{s - st}{t - t^2}X_{t} and E[Xs∣Xt=1]=s−stt−t2\mathbb{E}[X_{s} | X_{t} = 1] = \frac{s - st}{t - t^2}.
Key Concept
Conditional expectation in the Gaussian case
Explanation
The conditional expectation of one Gaussian random variable given another is linear, and we can use the covariance and variance to express it. When given a specific value for XtX_{t}, we can directly substitute it into the conditional expectation formula.
Consider the process Xt=μt+σBtt≥0. X_{t}=\mu t+\sigma B_{t} \quad t \geq 0 . for constants \mu \in \mathbb{R}, \sigma>0. Derive E[Xt]\mathbb{E}\left[X_{t}\right] and Cov⁡(Xt,Xs)\operatorname{Cov}\left(X_{t}, X_{s}\right) for some fixed t,s∈[0,1]t, s \in[0,1] and specify the joint distribution of XtX_{t} and XsX_{s}.
Solution by Steps
step 1
To derive E[Xt]\mathbb{E}[X_{t}], we use the linearity of expectation and the fact that BtB_t is a standard Brownian motion with E[Bt]=0\mathbb{E}[B_t] = 0
$$ \mathbb{E}[X_{t}] = \mathbb{E}[\mu t + \sigma B_{t}] = \mu t + \sigma \mathbb{E}[B_{t}] = \mu t $$
step 2
To find Cov⁡(Xt,Xs)\operatorname{Cov}(X_{t}, X_{s}), we use the properties of covariance and the fact that BtB_t has independent increments
$$ \operatorname{Cov}(X_{t}, X_{s}) = \operatorname{Cov}(\mu t + \sigma B_{t}, \mu s + \sigma B_{s}) = \sigma^2 \operatorname{Cov}(B_{t}, B_{s}) $$
step 3
Since BtB_t and BsB_s are standard Brownian motions, Cov⁡(Bt,Bs)=min⁡(t,s)\operatorname{Cov}(B_{t}, B_{s}) = \min(t, s)
$$ \operatorname{Cov}(X_{t}, X_{s}) = \sigma^2 \min(t, s) $$
step 4
The joint distribution of XtX_{t} and XsX_{s} is bivariate normal because any linear combination of normally distributed variables is also normally distributed
The mean vector is $\begin{bmatrix} \mu t \\ \mu s \end{bmatrix}$ and the covariance matrix is $\begin{bmatrix} \sigma^2 t &amp; \sigma^2 \min(t, s) \\ \sigma^2 \min(t, s) &amp; \sigma^2 s \end{bmatrix}$.
Answer
E[Xt]=μt\mathbb{E}[X_{t}] = \mu t, Cov⁡(Xt,Xs)=σ2min⁡(t,s)\operatorname{Cov}(X_{t}, X_{s}) = \sigma^2 \min(t, s), and the joint distribution of XtX_{t} and XsX_{s} is bivariate normal with mean vector [μtμs]\begin{bmatrix} \mu t \\ \mu s \end{bmatrix} and covariance matrix [σ2tamp;σ2min⁡(t,s)σ2min⁡(t,s)amp;σ2s]\begin{bmatrix} \sigma^2 t &amp; \sigma^2 \min(t, s) \\ \sigma^2 \min(t, s) &amp; \sigma^2 s \end{bmatrix}.
Key Concept
Expectation and covariance of a linear Brownian motion process
Explanation
The expectation of XtX_t is derived using the linearity of expectation, and the covariance is derived using the properties of covariance and the fact that Brownian motion has independent increments. The joint distribution is bivariate normal due to the properties of linear combinations of normal variables.
Consider the process Xt=μt+σBtt≥0. X_{t}=\mu t+\sigma B_{t} \quad t \geq 0 . for constants \mu \in \mathbb{R}, \sigma>0. Derive E[Xt]\mathbb{E}\left[X_{t}\right] and Cov⁡(Xt,Xs)\operatorname{Cov}\left(X_{t}, X_{s}\right) for some fixed t,s∈[0,1]t, s \in[0,1] and specify the joint distribution of XtX_{t} and XsX_{s}.
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