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You ask for the correlation function, defined for every $0\le s\le t\le u\le v$ by the formula $$ C(s,t;u,v)=E(Y_{s,t}Y_{u,v})-E(Y_{s,t})E(Y_{u,v}). $$ Let us first compute $E(Y_{s,t})$. For every $t\ge0$, introduce $$ M_t=\max\{B_s;0\le s\le t\}, $$ where $(B_t)$ is another standard Brownian motion, independent from $(W_t)$. Désiré André's reflection principle yields $$ P(M_t\ge x)=2P(B_t\ge x) $$ for every $x\ge0$, hence $E(M_t)=\sqrt{2t/\pi}$. This interests us because $Y_{s,t}$ is distributed like $W_s+M_{t-s}$, hence $$ E(Y_{s,t})=\sqrt{2(t-s)/\pi}. $$ To compute $E(Y_{s,t}Y_{u,v})$, let us use the decompositions $$Y_{s,t}=W_s+M_{t-s},\qquad Y_{u,v}=W_s+B_{t-s}+Z, $$ where $Z$ is independent on everything else. (And $Z=V_{u-t}+N_{v-u}$ where $V_{u-t}$ and $N_{v-u}$ are independent and independent from everything else, $V_{u-t}$ is distributed like $W_{u-t}$ and $N_{v-u}$ is distributed like $M_{v-u}$, but we will not use this.)

The fact that $W_s$ is centered and the independence properties given above yield $$ C(s,t;u,v)=E(W_s^2)+E(B_{t-s}M_{t-s})=s+E(B_{t-s}M_{t-s}). $$ The computation of $E(B_tM_t)$ is standard. One can use once again André's reflexion principle, which says that $$ P(M_t\ge x,B_t\in\mathrm{d}y)=g_t(\max(2x-y,y))\mathrm{d}y, $$ for every $x\ge0$, where $g_t$ is the centered Gaussian density of variance $t$. Now, $$ E(B_tM_t)=\int_0^{+\infty}\mathrm{d}x\int_{\mathbb{R}}yP(M_t\ge x,B_t\in\mathrm{d}y). $$ A (carefully executed) interversion of the order of integration yields $$ E(B_tM_t)=\int_0^{+\infty}y^2g_t(y)\mathrm{d}y=\frac12E(B_t^2)=\frac12t, $$ and finally, $$ \mathrm{Cov}(Y_{s,t},Y_{u,v})=\frac12(s+t). $$ Edit Somebody should check the numerics above. But the method is sound, and it gives also (I think) $$ E((Y_{s,t})^2)=4t-3s. $$

You ask for the correlation function, defined for every $0\le s\le t\le u\le v$ by the formula $$ C(s,t;u,v)=E(Y_{s,t}Y_{u,v})-E(Y_{s,t})E(Y_{u,v}). $$ One first computes $E(Y_{s,t})$. For every $t\ge0$, one introduces $$ M_t=\max\{B_s;0\le s\le t\}, $$ where $(B_t)$ is another standard Brownian motion, independent from $(W_t)$. For every $x\ge0$, Désiré André's reflection principle yields $$ P(M_t\ge x)=2P(B_t\ge x)=P(|B_t|\ge x). $$ One gets $E(M_t)=E(|B_t|)=\sqrt{2t/\pi}$. Since $Y_{s,t}$ is distributed like $W_s+M_{t-s}$, one deduces that $$ E(Y_{s,t})=\sqrt{2(t-s)/\pi}. $$ To compute $E(Y_{s,t}Y_{u,v})$, one can use the decompositions $$Y_{s,t}=W_s+M_{t-s},\qquad Y_{u,v}=W_s+B_{t-s}+Z, $$ where $Z$ is independent on everything else. (And $Z=V_{u-t}+N_{v-u}$ where $V_{u-t}$ and $N_{v-u}$ are independent and independent from everything else, $V_{u-t}$ is distributed like $W_{u-t}$ and $N_{v-u}$ is distributed like $M_{v-u}$, but one will not need this.)

The fact that $W_s$ is centered and the independence properties given above yield $$ C(s,t;u,v)=E(W_s^2)+E(B_{t-s}M_{t-s})=s+E(B_{t-s}M_{t-s}). $$ The computation of $E(B_tM_t)$ is standard. One can use once again André's reflexion principle, which says that $$ P(M_t\ge x,B_t\in\mathrm{d}y)=g_t(\max(2x-y,y))\mathrm{d}y, $$ for every $x\ge0$, where $g_t$ is the centered Gaussian density of variance $t$. Now, $$ E(B_tM_t)=\int_0^{+\infty}\mathrm{d}x\int_{\mathbb{R}}yP(M_t\ge x,B_t\in\mathrm{d}y). $$ A (carefully executed) interversion of the order of integration yields $$ E(B_tM_t)=\int_0^{+\infty}y^2g_t(y)\mathrm{d}y=\frac12E(B_t^2)=\frac12t, $$ and finally, $$ \mathrm{Cov}(Y_{s,t},Y_{u,v})=\frac12(s+t). $$ Special cases are $$ \mathrm{Cov}(Y_{s,t},B_{u})=\frac12(s+t),\quad \mathrm{Cov}(B_{t},Y_{u,v})=t, $$ and the same method yields $$ E((Y_{s,t})^2)=4t-3s. $$

You ask for the correlation function, defined for every $0\le s\le t\le u\le v$ by the formula $$ C(s,t;u,v)=E(Y_{s,t}Y_{u,v})-E(Y_{s,t})E(Y_{u,v}). $$ Let us first compute $E(Y_{s,t})$. For every $t\ge0$, introduce $$ M_t=\max\{B_s;0\le s\le t\}, $$ where $(B_t)$ is another standard Brownian motion, independent from $(W_t)$. By Désiré André's reflection principle, $P(M_t\ge x)=2P(B_t\ge x)$ for every $x\ge0$, hence $E(M_t)=\sqrt{2t/\pi}$. This interests us because $Y_{s,t}$ is distributed like $W_s+M_{t-s}$, hence $$ E(Y_{s,t})=\sqrt{2(t-s)/\pi}. $$ To compute $E(Y_{s,t}Y_{u,v})$, let us use the decompositions $$Y_{s,t}=W_s+M_{t-s},\qquad Y_{u,v}=W_s+B_{t-s}+Z, $$ where $Z$ is independent on everything else. (And $Z=V_{u-t}+N_{v-u}$ where $V_{u-t}$ and $N_{v-u}$ are independent and independent from everything else, $V_{u-t}$ is distributed like $W_{u-t}$ and $N_{v-u}$ is distributed like $M_{v-u}$, but we will not use this.)

The fact that $W_s$ is centered and the independence properties given above yield $$ C(s,t;u,v)=E(W_s^2)+E(B_{t-s}M_{t-s})=s+m(t-s),\quad m(t)=E(B_tM_t). $$ The computation of $m(t)$ is standard. One can use once again André's reflexion principle, which says that $$ P(M_t\ge x,B_t\in\mathrm{d}y)=[g_t(2x-y)\mathbf{1}_{y\le x}+g_t(y)\mathbf{1}_{y > x}]\mathrm{d}y, $$ where $g_t$ denotes the density of $B_t$. Now, $$ m(t)=\int_0^{+\infty}\mathrm{d}x\int_{\mathbb{R}}yP(M_t\ge x,B_t\in\mathrm{d}y). $$ An interversion of the order of integrals carefully executed yields $$ m(t)=\int_0^{+\infty}y^2g_t(y)\mathrm{d}y=\frac12E(B_t^2)=\frac12t, $$ and finally, $$ \mathrm{Cov}(Y_{s,t},Y_{u,v})=\frac12(s+t). $$ Edit Somebody should check the numerics above. But the method is sound, and it gives also (I think) $$ E((Y_{s,t})^2)=4t-3s. $$

You ask for the correlation function, defined for every $0\le s\le t\le u\le v$ by the formula $$ C(s,t;u,v)=E(Y_{s,t}Y_{u,v})-E(Y_{s,t})E(Y_{u,v}). $$ Let us first compute $E(Y_{s,t})$. For every $t\ge0$, introduce $$ M_t=\max\{B_s;0\le s\le t\}, $$ where $(B_t)$ is another standard Brownian motion, independent from $(W_t)$. Désiré André's reflection principle yields $$ P(M_t\ge x)=2P(B_t\ge x) $$ for every $x\ge0$, hence $E(M_t)=\sqrt{2t/\pi}$. This interests us because $Y_{s,t}$ is distributed like $W_s+M_{t-s}$, hence $$ E(Y_{s,t})=\sqrt{2(t-s)/\pi}. $$ To compute $E(Y_{s,t}Y_{u,v})$, let us use the decompositions $$Y_{s,t}=W_s+M_{t-s},\qquad Y_{u,v}=W_s+B_{t-s}+Z, $$ where $Z$ is independent on everything else. (And $Z=V_{u-t}+N_{v-u}$ where $V_{u-t}$ and $N_{v-u}$ are independent and independent from everything else, $V_{u-t}$ is distributed like $W_{u-t}$ and $N_{v-u}$ is distributed like $M_{v-u}$, but we will not use this.)

The fact that $W_s$ is centered and the independence properties given above yield $$ C(s,t;u,v)=E(W_s^2)+E(B_{t-s}M_{t-s})=s+E(B_{t-s}M_{t-s}). $$ The computation of $E(B_tM_t)$ is standard. One can use once again André's reflexion principle, which says that $$ P(M_t\ge x,B_t\in\mathrm{d}y)=g_t(\max(2x-y,y))\mathrm{d}y, $$ for every $x\ge0$, where $g_t$ is the centered Gaussian density of variance $t$. Now, $$ E(B_tM_t)=\int_0^{+\infty}\mathrm{d}x\int_{\mathbb{R}}yP(M_t\ge x,B_t\in\mathrm{d}y). $$ A (carefully executed) interversion of the order of integration yields $$ E(B_tM_t)=\int_0^{+\infty}y^2g_t(y)\mathrm{d}y=\frac12E(B_t^2)=\frac12t, $$ and finally, $$ \mathrm{Cov}(Y_{s,t},Y_{u,v})=\frac12(s+t). $$ Edit Somebody should check the numerics above. But the method is sound, and it gives also (I think) $$ E((Y_{s,t})^2)=4t-3s. $$

You ask for the correlation function, defined for every $0\le s\le t\le u\le v$ by the formula $$ C(s,t;u,v)=E(Y_{s,t}Y_{u,v})-E(Y_{s,t})E(Y_{u,v}). $$ Let us first compute $E(Y_{s,t})$. For every $t\ge0$, introduce $$ M_t=\max\{B_s;0\le s\le t\}, $$ where $(B_t)$ is another standard Brownian motion, independent from $(W_t)$. By Désiré André's reflection principle, $P(M_t\ge x)=2P(B_t\ge x)$ for every $x\ge0$, hence $E(M_t)=\sqrt{2t/\pi}$. This interests us because $Y_{s,t}$ is distributed like $W_s+M_{t-s}$, hence $$ E(Y_{s,t})=\sqrt{2(t-s)/\pi}. $$ To compute $E(Y_{s,t}Y_{u,v})$, let us use the decompositions $$Y_{s,t}=W_s+M_{t-s},\qquad Y_{u,v}=W_s+B_{t-s}+Z, $$ where $Z$ is independent on everything else. (And $Z=V_{u-t}+N_{v-u}$ where $V_{u-t}$ and $N_{v-u}$ are independent and independent from everything else, $V_{u-t}$ is distributed like $W_{u-t}$ and $N_{v-u}$ is distributed like $M_{v-u}$, but we will not use this.)

The fact that $W_s$ is centered and the independence properties given above yield $$ C(s,t;u,v)=E(W_s^2)+E(B_{t-s}M_{t-s})=s+m(t-s),\quad m(t)=E(B_tM_t). $$ The computation of $m(t)$ is standard. One can use once again André's reflexion principle, which says that $$ P(M_t\ge x,B_t\in\mathrm{d}y)=[g_t(2x-y)\mathbf{1}_{y\le x}+g_t(y)\mathbf{1}_{y > x}]\mathrm{d}y, $$ where $g_t$ denotes the density of $B_t$. Now, $$ m(t)=\int_0^{+\infty}\mathrm{d}x\int_{\mathbb{R}}yP(M_t\ge x,B_t\in\mathrm{d}y). $$ An interversion of the order of integrals carefully executed yields $$ m(t)=\int_0^{+\infty}y^2g_t(y)\mathrm{d}y=\frac12E(B_t^2)=\frac12t, $$ and finally, $$ \mathrm{Cov}(Y_{s,t},Y_{u,v})=\frac12(s+t). $$

You ask for the correlation function, defined for every $0\le s\le t\le u\le v$ by the formula $$ C(s,t;u,v)=E(Y_{s,t}Y_{u,v})-E(Y_{s,t})E(Y_{u,v}). $$ Let us first compute $E(Y_{s,t})$. For every $t\ge0$, introduce $$ M_t=\max\{B_s;0\le s\le t\}, $$ where $(B_t)$ is another standard Brownian motion, independent from $(W_t)$. By Désiré André's reflection principle, $P(M_t\ge x)=2P(B_t\ge x)$ for every $x\ge0$, hence $E(M_t)=\sqrt{2t/\pi}$. This interests us because $Y_{s,t}$ is distributed like $W_s+M_{t-s}$, hence $$ E(Y_{s,t})=\sqrt{2(t-s)/\pi}. $$ To compute $E(Y_{s,t}Y_{u,v})$, let us use the decompositions $$Y_{s,t}=W_s+M_{t-s},\qquad Y_{u,v}=W_s+B_{t-s}+Z, $$ where $Z$ is independent on everything else. (And $Z=V_{u-t}+N_{v-u}$ where $V_{u-t}$ and $N_{v-u}$ are independent and independent from everything else, $V_{u-t}$ is distributed like $W_{u-t}$ and $N_{v-u}$ is distributed like $M_{v-u}$, but we will not use this.)

The fact that $W_s$ is centered and the independence properties given above yield $$ C(s,t;u,v)=E(W_s^2)+E(B_{t-s}M_{t-s})=s+m(t-s),\quad m(t)=E(B_tM_t). $$ The computation of $m(t)$ is standard. One can use once again André's reflexion principle, which says that $$ P(M_t\ge x,B_t\in\mathrm{d}y)=[g_t(2x-y)\mathbf{1}_{y\le x}+g_t(y)\mathbf{1}_{y > x}]\mathrm{d}y, $$ where $g_t$ denotes the density of $B_t$. Now, $$ m(t)=\int_0^{+\infty}\mathrm{d}x\int_{\mathbb{R}}yP(M_t\ge x,B_t\in\mathrm{d}y). $$ An interversion of the order of integrals carefully executed yields $$ m(t)=\int_0^{+\infty}y^2g_t(y)\mathrm{d}y=\frac12E(B_t^2)=\frac12t, $$ and finally, $$ \mathrm{Cov}(Y_{s,t},Y_{u,v})=\frac12(s+t). $$ Edit Somebody should check the numerics above. But the method is sound, and it gives also (I think) $$ E((Y_{s,t})^2)=4t-3s. $$